# On endomorphisms of groups of order 36/36. Jarku ruhmade endomorfismidest.

1. INTRODUCTIONIt is well known that all endomorphisms of an Abelian group form a ring and many of their properties can be characterized by this ring. An excellent overview of the present situation in the theory of endomorphism rings of Abelian groups is given by Krylov et al. [3]. All endomorphisms of an arbitrary group form only a semigroup. The theory of endomorphism semigroups of groups is quite modestly developed. In a number of our papers we have made efforts to describe some classes of groups that are determined by their endomorphism semigroups in the class of all groups. Let G be a group. If for each group H such that the semigroups End(G) and End(H) are isomorphic implies an isomorphism between G and H, then we say that the group G is determined by its endomorphism semigroup in the class of all groups. Examples of such groups are finite Abelian groups ([4], Theorem 4.2), generalized quaternion groups ([5], Corollary 1), torsion-free divisible Abelian groups ([6], Theorem 1), etc.

We know a complete answer to this problem for finite groups of order less than 36. The alternating group [A.sub.4] (also called the tetrahedral group) and the binary tetrahedral group B = <a, b | [b.sup.3] = 1, aba = bab> are the only groups of order less than 32 that are not determined by their endomorphism semigroups in the class of all groups [12]. These two groups are non-isomorphic, but their endomorphism semigroups are isomorphic. We have proved that each group of order 32 is determined by its endomorphism semigroup in the class of all groups: it has partly been made in published papers [13,14] and partly in papers to be published. The groups of orders 33 and 35 are cyclic, and, therefore, are determined by their endomorphism semigroups in the class of all groups ([4], Theorem 4.2). There exist two non-isomorphic groups of order 34: the cyclic group of order 34 and the dihedral group of order 34. Both are determined by their endomorphism semigroups in the class of all groups ([4], Theorem 4.2 and [10], Section 5).

In this paper, we present a solution to the problem whose groups of order 36 are determined by their endomorphism semigroups. The group theoretical computer algebra system GAP contains the 'Small Groups Library', which provides access to descriptions of all groups of order 36 ([17]). There exist exactly 14 non-isomorphic groups of order 36. Throughout this paper, let us denote these groups by [G.sub.1], [G.sub.2], ..., [G.sub.14], respectively. The last three groups among them are

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where [C.sub.3] is the cyclic group of order 3 and A4 is the alternating group of order 12 (the tetrahedral group). In this paper, the following theorem is proved:

Theorem 1.1 (Main theorem). The following statements hold for a group G:

(1) if the endomorphism semigroups of G and [G.sub.i], i [member of] {1, 2, ..., 11} are isomorphic, then G and [G.sub.i] are isomorphic;

(2) the endomorphism semigroups of G and [G.sub.12] are isomorphic if and only if G = [G.sub.12] or

G = <a, b, c | [c.sup.9] = [a.sup.4] = 1, [a.sup.2] = [b.sup.2], [b.sup.-1] ab = [a.sup.-1], [c.sup.-1] bc = a, [c.sup.-1] ac = ab>;

(3) the endomorphism semigroups of G and [G.sub.13] are isomorphic if and only if G = [G.sub.13] or G = [C.sub.3] x B, where B = <a, b | [b.sup.3] = 1 , aba = bab> is the binary tetrahedral group;

(4) the endomorphism semigroups of G and [G.sub.14] are isomorphic if and only if G = [G.sub.14] or

G = {a, b, c, d | [c.sup.4] = [a.sup.3] = [b.sup.3] = [d.sup.3] = 1, ab = bad, [c.sup.-1] ac = b, [c.sup.-1] bc = [a.sup.-1], cd = dc, ad = da, bd = db>.

We shall use the following notations:

G--a group;

Z(G)--the centre of a group G;

G'--the derived subgroup of G;

[a, b] = [a.sup.-1] [b.sup.-1] ab (a, b [member of] G);

[C.sub.G](a)--the centralizer of a in G;

End(G)--the endomorphism semigroup of G;

[C.sub.k]--the cyclic group of order k;

[A.sub.4]--the alternating group of order 12 (the tetrahedral group);

[D.sub.n] = <a, b | [b.sup.2] = [a.sup.k] = 1, [b.sup.-1] ab = [a.sup.-1]>--the dihedral group of order n = 2k;

B = <a, b | [b.sup.3] = 1, aba = bab>--the binary tetrahedral group;

[Z.sub.k]--the residue class ring Z/kZ;

[Z.sub.k][x]--the polynomial ring over [Z.sub.k];

<K, ..., g, ...>--the subgroup generated by subsets K,... and elements g, ...;

[??]--the inner automorphism of G, generated by an element g [member of] G;

I(G)--the set of all idempotents of End(G);

K(x) = {z [member of] End(G) | zx = xz = z};

J(x) = {z [member of] End(G) | zx = xz = 0};

V(x) = {z [member of] Aut(G) | zx = x};

D(x) = {z [member of] Aut(G) | zx = xz = x};

H(x) = {z [member of] End(G) | xz = z, zx = 0};

[x] = {z [member of] I(G) | xz = z, zx = x}, x [member of] I(G);

G = A [lambda] B - G is a semidirect product of an invariant subgroup A and a subgroup B.

The sets K(x), V(x), D(x), and J(x) are subsemigroups of End(G); however, V(x) and D(x) are subgroups of Aut(G). We shall write the mapping right from the element on which it acts.

2. PRELIMINARIES

For the convenience of the reader, let us recall some known facts that will be used in the proofs of our main results.

Lemma 2.1. If x [member of] I(G), then G = Ker x [lambda] Im x and Im x = {g [member of] G | gx = g}.

Lemma 2.2. If x [member of] I(G), then [x] = {y [member of] I(G) | Ker x = Ker y}.

Lemma 2.3. If x [member of] I(G), then

K(x) = {y [member of] End(G) | (Imx)y [subset] Imx, (Kerx)y = <1>}

and K(x) is a subsemigroup with the unity x of End(G), which is canonically isomorphic to End(Imx). Under this isomorphism element y of K(x) corresponds to its restriction onto the subgroup Imx of G.

Lemma 2.4. If x [member of] I(G), then

H (x) = {y [member of] End(G) | (Im x)y [subset] Ker x, (Ker x)y = <1>}.

Lemma 2.5. If x [member of] I(G), then

J(x) = {z [member of] End(G) | (Imx)z = <1>, (Ker x)z [subset] Ker x}.

Lemma 2.6. If x [member of] I(G), then

D(x) = {y [member of] Aut(G) [[absolute value of (y)].sub.Imx] = 1[|.sub.Imx], (Ker x)y C Ker x}.

Lemma 2.7. Ifz [member of] End(G) and Im z is Abelian, then [??] [member of] V(z) for each g [member of] G.

We omit the proofs of these lemmas because these are straightforward corollaries from the definitions.

Lemma 2.8 ([4], Theorem 4.2). Every finite Abelian group is determined by its endomorphism semigroup in the class of all groups.

Lemma 2.9 ([4], Theorem 1.13). If G and H are groups such that their endomorphism semigroups are isomorphic and G splits into a direct product G = [G.sub.1] x [G.sub.2] of its subgroups [G.sub.1] and [G.sub.2], then H splits into a direct product H = [H.sub.1] x [H.sub.2] of its subgroups [H.sub.1] and [H.sub.2] such that End([G.sub.1]) [congruent to] End([H.sub.1]) and End([G.sub.2]) [congruent to] End([H.sub.2]).

From here follow Lemmas 2.10-2.13.

Lemma 2.10. If groups [G.sub.1] and [G.sub.2] are determined by their endomorphism semigroups in the class of all groups, then so is their direct product [G.sub.1] x [G.sub.2].

Lemma 2.11 ([10], Section 5). The dihedral group [D.sub.n] is determined by its endomorphism semigroup in the class of all groups.

Lemma 2.12 ([9], Theorem, Lemmas 4.5-4.8). Let

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where p is a prime, p > 2, and let [G.sup.*] be another group such that the endomorphism semigroups of G and [G.sup.*] are isomorphic. Assume that x is the projection of G onto its subgroup <b> and [x.sup.*] corresponds to x under the isomorphism End(G) [congruent to] End([G.sup.*]). Then G and [G.sup.*] are isomorphic and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where Im [x.sup.*] = <d>, Ker [x.sup.*] = <c>, and <r> = <[r.sup.*]> in the group of units of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Lemma 2.13 ([2], Theorem 2.1 and Lemma 2.2). If G is a group such that G/Z(G) is Abelian, then

[g, hk] = [g, h] x [g, k], [hk, g] = [h, g] x [k, g], [(gh).sup.m] = [g.sup.m][h.sup.m][[h, g].sup.m(m-1)/2], [[g, h].sup.m] = [[g.sup.m], h] = [g, [h.sup.m]]

for each g, k, h [member of] G and positive integer m.

3. GROUPS OF ORDER 36

The group theoretical computer algebra system GAP provides access to descriptions of small order groups [17]. At present, the library of small order groups contains the groups of order at most 2000, except for order 1024 (423 164 062 groups).

There are 14 pairwise non-isomorphic groups of order 36. Following [17], they are

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One of the main ideas to prove the main theorem is to use Lemmas 2.1-2.13. Therefore, we present some of the groups [G.sub.i] (1 [less than or equal to] i [less than or equal to] 14) in the form suitable for these lemmas.

(1) The group G6. It follows from the determining relations of G6:

ab = ba, bd = db; [a.sup.-1] dad = 1 [??] [a.sup.-1] da = [d.sup.-1],

and, therefore,

[G.sub.6] = <b> x (<d> [lambda] <a>) [congruent to] [C.sub.3] x ([C.sub.3] [lambda] [C.sub.4]). (3.1)

(2) The group [G.sub.7]. It follows from the determining relations of [G.sub.7]:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Therefore,

[G.sub.7] = <b> x (<c> [lambda] <a>) [congruent to] [C.sub.2] x [D.sub.18]. (3.2)

(3) The group [G.sub.8]. Rewriting the determining relations of [G.sub.8], we get

[a.sup.-1] da = [d.sup.-1], [b.sup.-1] cb = [c.sup.-1], ab = ba, ac = ca, bd = db, dc = cd.

Hence

[G.sub.8] = (<d> [lambda] <a>) x (<c> [lambda] <b>) [congruent to] [D.sub.6] x [D.sub.6]. (3.3)

(4) The group [G.sub.9]. It follows from the determining relations of [G.sub.9]:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (3.4)

(5) The group [G.sub.10]. It follows from the determining relations of [G.sub.10]:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Hence

<c> [lambda] <a> [congruent to] <d> [lambda] <a> [congruent to] [C.sub.3] [lambda] [C.sub.4]

and

[G.sub.10] = (<c> x <d>) [lambda] <a> = <c> [lambda] (<d> [lambda] <a>) = <d> [lambda] (<c> [lambda] <a>) [congruent to] ([C.sub.3] x [C.sub.3]) [lambda] [C.sub.4].

(6) The group [G.sub.11]. It follows from the determining relations of [G.sub.11]:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

(7) The group G12. It follows from the determining relations of G12:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Denote the elements a, c, and [a.sup.-1] ca by c, a, and b, respectively. Then we get

[G.sub.12] = <a, b, c | [c.sup.9] = [b.sup.2] = [a.sup.2] = 1, ab = ba, [c.sup.-1] ac = b, [c.sup.-1] bc = ab> = (<a> x <b>) [lambda] <c> [congruent to] ([C.sub.2] x [C.sub.2]) [lambda] [C.sub.9].

(8) The group G13. It follows from the determining relations of G13:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Denote the elements [a.sup.-1] ca, c, a, and b by b, a, c, and d, respectively. Then

<a, b, c> = <a, b, c | [c.sup.3] = [b.sup.2] = [a.sup.2] = 1, ab = ba, [c.sup.-1] ac = b, [c.sup.-1] bc = ab> = [A.sub.4], [G.sub.13] = <d> x ((<a> x <b>) [lambda] <c>) [congruent to] [C.sub.3] x [A.sub.4].

(9) The group G14. It follows from the determining relations of G14:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Denote the elements c, [a.sup.-1] ca, and a by a, b, and c, respectively. Then

[G.sub.14] = <a, b, c | [c.sup.4] = [a.sup.3] = [b.sup.3] = 1, ab = ba, [c.sup.-1] ac = b, [c.sup.-1] bc = [a.sup.-1]> = (<a> x <b>) [lambda] <c> [congruent to] ([C.sub.3] x [C.sub.3]) [lambda][C.sub.4].

By Lemmas 2.8-2.12 and Eqs (3.1)-(3.4), the groups [G.sub.1]-[G.sub.9] are determined by their endomorphism semigroups in the class of all groups.

4. ON ENDOMORPHISMS OF [G.sub.10] AND [G.sub.11]

In this section, we shall prove the following theorem.

Theorem 4.1. The groups

[G.sub.10] = <a, c, d | [a.sup.4] = [c.sup.3] = [d.sup.3] = 1, [a.sup.-1] cac = 1, [a.sup.-1] dad = 1, [d.sup.-1][c.sup.-1]dc = 1, [([a.sup.-1] c).sup.2] [a.sup.-2] = 1, [c.sup.-1] d[a.sup.-1] d[c.sup.-1] a = 1>

and

[G.sub.11] = <a, b, c, d | [a.sup.2] = [b.sup.2] = [c.sup.3] = [d.sup.3] = 1, [(ac).sup.2] = 1, [(ad).sup.2] = 1, [(ba).sup.2] = 1, [c.sup.-1] bcb = 1, [d.sup.-1] bdb = 1, [d.sup.-1][c.sup.-1] dc = 1>

are determined by their endomorphism semigroups in the class of all groups.

Proof. Let G be a group and [G.sub.1], [G.sub.2], K be subgroups of G such that G decomposes as follows:

G = ([G.sub.1] x [G.sub.2]) [lambda] K = [G.sub.1] [lambda] ([G.sub.2] [lambda] K) = [G.sub.2] [lambda] ([G.sub.1] [lambda]K). (4.1)

Denote by x, [x.sub.1], and [x.sub.2] the projections of G onto its subgroups K, [G.sub.1] [lambda] K, and [G.sub.2] [lambda] K, respectively. Then

Im x = K, Im [x.sub.1] = [G.sub.1] [lambda] K, Im [x.sub.2] = [G.sub.2] [lambda] K, Ker x = [G.sub.1] x [G.sub.2], Ker[x.sub.1] = [G.sub.2], Ker [x.sub.2] = [G.sub.1].

Assume that [G.sup.*] is another group such that the endomorphism semigroups of G and [G.sup.*] are isomorphic and [x.sup.*], [x.sup.*.sub.1], and [x.sup.*.sub.2] correspond to x, [x.sub.1], and [x.sub.2] in this isomorphism. Under these assumptions the group [G.sup.*] decomposes similarly to (4.1) [7, Theorems 2.1 and 3.1], i.e.,

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Let us apply this result to the following group G = G(n) :

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Hence

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By Lemma 2.3,

End(Im [x.sub.i]) [congruent to] K ([x.sub.i]) [congruent to] K([x.sup.*.sub.i]) [congruent to] End(Im [x.sup.*.sub.i]). (42)

From Lemma 2.12 (in the case of p = 3, n = 1, v = [2.sup.n]) and from (4.2) we have that

[K.sup.*] = Im [x.sup.*] = <[a.sub.0]>,

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Therefore,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Hence [G.sup.*] [congruent to] G(n) and the group G(n) is determined by its endomorphism semigroup in the class of all groups.

By the description of the groups [G.sub.10] and [G.sub.11], we have

[G.sub.10] [congruent to] G(2) and [G.sub.11] [congruent to] [C.sub.2] x G(1).

We conclude from Lemmas 2.8 and 2.9 that the groups [G.sub.10] and [G.sub.11] are determined by their endomorphism semigroups in the class of all groups. The theorem is proved.

Hence part (1) of Theorem 1.1 is proved.

5. ON ENDOMORPHISMS OF [G.sub.12]

Let us consider the group

[G.sub.12] = <a, b, c | [c.sup.9] = [b.sup.2] = [a.sup.2] = 1, ab = ba, [c.sup.-1] ac = b, [c.sup.-1] bc = ab> = (<a> x <b>) [lambda] <c> [congruent to] ([C.sub.2] x [C.sub.2]) [lambda] [C.sub.9].

In this section, we shall prove the following theorem:

Theorem 5.1. The endomorphism semigroup of a group G is isomorphic to the endomorphism semigroup of the group [G.sub.12] if and only if G = [G.sub.12] or

G = <a, b, c | [c.sup.9] = [a.sup.4] = 1, [a.sup.2] = [b.sup.2], [b.sup.-1] ab = [a.sup.-1], [c.sup.-1] bc = a, [c.sup.-1] ac = ab>.

Proof. The group [G.sub.12] is a Schmidt group, i.e., a non-nilpotent finite group in which each proper subgroup is nilpotent. The structure of the Schmidt groups is well known (see [15,16]). In [8] and [11], the endomorphisms of Schmidt groups are characterized. To prove the theorem, we present a summary of necessary results on Schmidt groups.

Each Schmidt group G can be described by three parameters p, q, and v, where p and q are different primes and v is a natural number, v [greater than or equal to] 1. A Schmidt group is not uniquely determined by its parameters. Fix parameters p, q, and v and denote by l the class of all Schmidt groups that have these parameters. There exist a group [G.sub.max] of the maximal order and a group Gmin of the minimal order in the class l. The groups [G.sub.max] and [G.sub.min] are uniquely determined up to the isomorphism. A group G belongs to l if and only if it is isomorphic to the factor-group [G.sub.max]/M, where M is a subgroup of the second derived group [G".sub.max] of [G.sub.max]. Clearly, [G.sub.min] [congruent to] [G.sub.max]/[G".sub.max]. Let us give the description of [G.sub.min].

Assume that [psi](x) is an arbitrary irreducible normalized divisor of the polynomial

[x.sup.q] - 1/x - 1 = [x.sup.q-1] + [x.sup.q-2] + ... + x + 1 [member of] [Z.sub.p] [x].

Suppose that the degree of [psi] is u. Denote by [[bar.Z].sub.p][x] the residue class ring [Z.sub.p][x]/[psi](x)[Z.sub.p][x]. Then

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and the composition rule in [G.sub.min] is

(i; f (x)) x (j; g(x)) = (i + j; f (x) + [x.sup.i] x g(x))

([15], Proposition 7). If u is odd, then [G.sub.max] = [G.sub.min] and all Schmidt groups with parameters p, q, and v are isomorphic. If u is even, say u = 2t, then [absolute value of ([G.sub.max])] = [q.sup.v][p.sup.u+t] and [G.sub.max] can be given by generators and generating relations in which the coefficients of the polynomial [psi](x) are used ([16], Proposition 3).

Fix now the parameters p, q, and v as follows:

p = 2, q = 3, v = 2.

Then

[x.sup.q] - 1/x - 1 = [x.sup.3] - 1/x - 1 = [x.sup.2] + x + 1 [member of] [Z.sub.2] [x].

Since the polynomial [x.sup.2] + x + 1 is irreducible in [Z.sub.2][x], we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

It follows that

l = {[G.sub.min], [G.sub.max]}. (5.1)

The group [G.sub.min] consists of pairs (i; f (x)), where i [member of] [Z.sub.9] and f (x) [member of] [[bar.Z].sub.2][x]. Denote c = (1; 0), a = (0; x), and b = (0; 1). Then [G.sub.min] can be given as follows:

[G.sub.min] = <a, b, c | [c.sup.9] = [b.sup.2] = [a.sup.2] = 1, ab = ba, [c.sup.-1] ac = b, [c.sup.-1]bc = ab> = (<a> x <b>) [lambda] <c> [congruent to] ([C.sub.2] x [C.sub.2]) [lambda] [C.sub.9].

Therefore,

[G.sub.min] = [G.sub.12]. (5.2)

The group [G.sub.max] is given as follows ([16, Proposition 3]):

[G.sub.max] = <a, b, c | [c.sup.9] = [a.sup.4] = 1, [a.sup.2] = [b.sup.2], [b.sup.-1] ab = [a.sup.-1], [c.sup.-1]bc = a, [c.sup.-1]ac = ab>. (5.3)

The endomorphism semigroups of [G.sub.max] and [G.sub.min] are isomorphic ([8], Theorem 4.4):

End ([G.sub.max]) = End([G.sub.min]). (5.4)

Let [G.sup.*] be a group such that End([G.sup.*]) [congruent to] End([G.sub.12]) [congruent to] End([G.sub.min]). By [11], Theorem 3.2, the group [G.sup.*] is also a Schmidt group with the same parameters as [G.sub.min], i. e., [G.sup.*] [member of] S. Therefore, [G.sup.*] [congruent to] [G.sub.min] [congruent to] G(12) or [G.sup.*] [congruent to] [G.sub.max]. Isomorphism (5.4) implies the statement of the theorem. The theorem is proved and so is part (2) of Theorem 1.1.

6. ON ENDOMORPHISMS OF [G.sub.13]

Let us consider the group

[G.sub.13] = <d> x ((<a> x <b>) [lambda] <c>) [congruent to] [C.sub.3] x [A.sub.4].

In this section, we shall find all groups G such that End(G) [congruent to] End([G.sub.13]).

Similarly to the previous section, there exist only two Schmidt groups with the parameters p = 2, q = 3, and v = 1. They are

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Taking in the presentation of the binary tetrahedral group B = <[alpha], [beta] | [[beta].sup.3] = 1, [alpha][beta][alpha] = [beta][alpha][beta]> new generators c = [beta], b = [alpha][[beta].sup.-1], a = [alpha][beta][alpha], we see that B coincides with [G.sub.max]. Since the second commutator of B is B" = <[a.sup.2]> and [G.sub.min] [congruent to] [G.sub.max]/[G".sub.max], we have

[A.sub.4] = [G.sub.min] [congruent to] [G.sub.max]/[G".sub.max] = B/B" = B/<[a.sup.2]>.

Let us identify [A.sub.4] = B/<[a.sup.2]> and denote the elements of the factor-group B/<[a.sup.2]> by [bar.g] = g x <[a.sup.2]>, g [member of] B. Then

B = <a, b, c | [c.sup.3] = [a.sup.4] = 1, [a.sup.2] = [b.sup.2], [b.sup.-1] ab = [a.sup.-1], [c.sup.-1] ac = b, [c.sup.-1] bc = ab> = <a, b> [lambda] <c>. (6.1)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (6.2)

Lemma 6.1. Let [G.sup.*] be a group. Then End([G.sup.*]) [congruent to] End([A.sub.4]) if and only if [G.sup.*] = [A.sub.4] or [G.sup.*] = B. The map T : End (B) [right arrow] End(B/B") = End([A.sub.4]) defined by

[tau]T = [bar.[tau]], [tau] [member of] End (B), [bar.g][bar.[tau]] = [bar.g[tau]], g [member of] B,

is the isomorphism of semigroups.

Proof. The proof of the first statement is the same as in the last part of the proof of Theorem 5.1. The second statement of the lemma follows from isomorphism (5.4), which is described in ([8], Theorem 4.4). The lemma is proved.

Lemma 6.2. End([C.sub.3] x B) [congruent to] End([C.sub.3] x [A.sub.4]).

Proof. Assume that [C.sub.3] = <d>. By (6.1) and (6.2),

[C.sub.3] x B = <d> x (<a, b> [lambda] <c>), (6.3)

[C.sub.3] x [A.sub.4] = ([C.sub.3] x B)/<[a.sup.2]> = <d> x (<[bar.a], [bar.b]> [lambda] <[bar.c]>), (6.4)

where [bar.d] = d x <[a.sup.2]> is identified with d.

The groups B and [A.sub.4] satisfy the following two properties:

(1) each 3-element of [A.sub.4] has the form [bar.g], where g is a 3-element of B and the numbers of 3-elements of groups [A.sub.4] and B coincide;

(2) if g is a 3-element of B and g [not equal to] 1, then <g> [congruent to] <[bar.g]> [congruent to] [C.sub.3] and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

In view of (6.3) and (6.4), each endomorphism T of [C.sub.3] x B has the form

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where i, k [member of] [Z.sub.3], [tau] [member of] End(B), and g is a 3-element of B such that it commutes with C[tau], i.e. (a) if C[tau] = 1, then g is an arbitrary 3-element of B; (b) if C[tau] [not equal to] 1, then, by property (2), g = [(c[tau]).sup.j], where j [member of] [Z.sub.3]. Similarly, each endomorphism [bar.T] of [C.sub.3] x [A.sub.4] has the form

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where i, k [member of] [Z.sub.3], t [member of] End(B), and g is a 3-element of B such that it commutes with c[tau]. It follows that

[absolute value of (End([C.sub.3] x B))] = [absolute value of (End([C.sub.3] x [A.sub.4]))], [bar.g][bar.T] = [bar.gT] for each g [member of] [C.sub.3] x B

and it is easy to see that the map T [right arrow] [bar.T] gives an isomorphism End([C.sub.3] x B) [congruent to] End([C.sub.3] x [A.sub.4]). The lemma is proved.

Theorem 6.1. The endomorphism semigroup of a group G is isomorphic to the endomorphism semigroup of the group [G.sub.13] = [C.sub.3] x [A.sub.4] if and only if G = [C.sub.3] x [A.sub.4] or G = [C.sub.3] x B.

Proof. Let G be a group such that

End(G) [congruent to] End([G.sub.13]) [congruent to] End([C.sub.3] x [A.sub.4]).

Since [G.sub.13] is finite, G is finite, too ([1], Theorem 2). By Lemma 2.12, the group G splits into a direct product G = C x D such that

End(C) [congruent to] End([C.sub.3]), End(D) [congruent to] End([A.sub.4]).

By Lemmas 2.8 and 6.1, C [congruent to] [C.sub.3] and D [congruent to] [A.sub.4] or D [congruent to] B. It follows that G [congruent to] [C.sub.3] x [A.sub.4] or G [congruent to] [C.sub.3] x B. By Lemma 6.2, the statement of the theorem is true. The theorem is proved and so is part 3 of Theorem 1.1.

7. ON ENDOMORPHISMS OF [G.sub.14]

In this section, we shall consider the group

[G.sub.14] = <a, b, c | [c.sup.4] = [a.sup.3] = [b.sup.3] = 1, ab = ba, [c.sup.-1]ac = b, [c.sup.-1]bc = [a.sup.-1]> = (<a> x <b>) [lambda] <c> [congruent to] ([C.sub.3] x [C.sub.3]) [lambda] [C.sub.4]. (7.1)

Our aim is to find all groups G such that End(G) [congruent to] End([G.sub.14]).

Lemma 7.1. The endomorphisms of [G.sub.14] are the zero-endomorphism and the following maps :

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]; (7.2)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (7.3)

The maps (7.2) and (7.3), where (k, l) [not equal to] (0, 0), are automorphisms and

[absolute value of (End([G.sub.14]))] = 172, [absolute value of (Aut([G.sub.14]))] = 144.

Each endomorphism of [G.sub.14] is induced by the images of the generators, and therefore, to prove the lemma, it is necessary to find conditions under which these images preserve the defining relations of [G.sub.14]. These easy calculations are omitted.

Consider the group

[G.sub.0] = <a, b, c, d | [c.sup.4] = [a.sup.3] = [b.sup.3] = [d.sup.3] = 1, ab = bad, [c.sup.-1]ac = b, [c.sup.-1]bc = [a.sup.-1], cd = dc, ad = da, bd = db>.

Clearly, Z([G.sub.0]) = <d> and

[G.sub.0]/<d> = [G.sub.0]/Z([G.sub.0]) = [G.sub.14]. (7.4)

Lemma 7.2. End([G.sub.0]) [congruent to] End([G.sub.14]).

Proof. Immediate calculations show that the endomorphisms of G0 are the zero-endomorphism and the following maps:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]; (7.5)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]; (7.6)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (7.7)

If [tau] [member of] End([G.sub.0]), then the map [bar.[tau]] : [G.sub.0]/Z([G.sub.0]) [right arrow] [G.sub.0]/Z([G.sub.0]), defined by

(g x Z([G.sub.0])[bar.[tau]] = (g[tau]) x Z([G.sub.0]), g [member of] [G.sub.0],

is an endomorphism of [G.sub.0]/Z([G.sub.0]). In view of Lemma 7.1, isomorphism (7.4) and endomorphisms (7.5)-(7.7), the map

T : End([G.sub.0]) [right arrow] End([G.sub.0]/Z([G.sub.0])), [tau]T = [bar.[tau]], [tau] [member of] End([G.sub.0])

is bijective. Since ([[tau].sub.1][[tau].sub.2])T = ([[tau].sub.1]T)([[tau].sub.2]T), the map T is an isomorphism. Therefore, End([G.sub.0]) = End([G.sub.0]/Z([G.sub.0])) = End([G.sub.14]). The lemma is proved.

By (7.1), we have [G.sub.14] = (<a> x <b>) [lambda] <c>. Denote by x the projection of [G.sub.14] onto its subgroup <c>. Then x [member of] I([G.sub.14]) and

Kerx = <a> x <b> [congruent to] [C.sub.3] x [C.sub.3], Imx = <c> [congruent to] [C.sub.4]. (7.8)

Lemma 7.3. The projection x given by (7.8) satisfies the following properties:

[1.sup.0] K(x) [congruent to] End([C.sub.4]);

[2.sup.0] H(x) = {0};

[3.sup.0] J(x) = {0};

[4.sup.0] D(x) [congruent to] [C.sub.8];

[5.sup.0] [absolute value of ([x])] = 9;

[6.sup.0] y [member of] [x], y [not equal to] x [??] K(x) [intersection] K(y) = {0};

[7.sup.0] V (x) has a subgroup isomorphic to [G.sub.14];

[8.sup.0] V(x) = A [lambda] D(x), where A [congruent to] [C.sub.3] x [C.sub.3].

Proof. By Lemma 2.3, K(x) [congruent to] End(Imx) [congruent to] End([C.sub.4]) and property [1.sup.0] is true. Lemma 2.4 and (7.8) imply property [2.sup.0]. Lemma 2.5 and (7.8) imply that an endomorphism z of [G.sub.14] belongs to J(x) if and only if cz = 1 and <a, b>z [subset] <a, b>, i.e., by Lemma 7.1, z = 0. Hence property [3.sup.0] holds.

By Lemma 2.6, D(x) consists of automorphisms y of [G.sub.14] such that

<a, b> y = <a, b>, cy = c.

In view of Lemma 7.1,

D(x) = {y | cy = c, by = [b.sup.l] [a.sup.-k], ay = [b.sup.k][a.sup.l]; k, l [member of] [Z.sub.3], (k, l) [not equal to] (0, 0)}.

Therefore, [absolute value of (D(x))] = 8. Choose z [member of] D(x) as follows:

cz = c, bz = b[a.sup.-1], az = ba. (7.9)

Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

i.e., z is an element of order 8, and, therefore, D(x) = <z> [congruent to] [C.sub.8] and property [4.sup.0] holds.

Using Lemma 7.1, it is easy to check that I([G.sub.14]) consists of 0, 1 and the maps

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Since x = [y.sub.0,0] and [y.sub.t,u]x = x, x[y.sub.t,u] = [y.sub.t,u], we have

[x] = {[y.sub.t,u] | t, u [member of] [Z.sub.3]}, [absolute value of ([x])] = 9.

Choose [y.sub.t,u] [member of] [x], [y.sup.t,u] [not equal to] x,i.e., (t, u) [not equal to] (0, 0). Since Imx = <c> and Im[y.sub.t,u] = <c[b.sup.t][a.sup.u]> are Sylow 2-subgroups of [G.sub.14], there exists g [member of] [G.sub.14] such that Im [y.sub.t,u] = <[g.sup.-1]cg>. We have Imx [intersection] Im[y.sub.t,u] = (1), because [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. It follows from here and Lemma 2.3 that K(x) [intersection] K([y.sub.t,u]) = {0}. Properties [5.sup.0] and [6.sup.0] are proved.

By Lemma 2.7, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since Z([G.sub.14]) = {1}, we have [G.sub.14] [congruent to] [[??.sup.14] [subset] V(x) and property [7.sup.0] is true.

Since an automorphism y of [G.sub.14] belongs to V (x) if and only if [g.sup.-1] x gy [member of] Ker x for each g [member of] G, Lemma 7.1 implies that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

i. e.,

[absolute value of (V (x))] = 8 x 9 = 72

and the Sylow 3-subgroup of V(x) is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

The Sylow 2-subgroup of V(x) is D(x) = <z> [congruent to] [C.sub.8], where z is given by (7.9). We have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

i.e., A [??] V(x) and V(x) = A [lambda] D(x). Property [8.sup.0] is true. The lemma is proved.

Lemma 7.4. If G is a finite group and there exists x [member of] I(G) such that it satisfies properties [1.sup.0] - [8.sup.0] of Lemma 7.3, then G is isomorphic to [G.sub.14] or [G.sub.0].

Proof. Let G be a finite group such that there exists x [member of] I(G) that satisfies properties [1.sup.0]-[8.sup.0] of Lemma 7.3.

By Lemma 2.1, we have G = Kerx [lambda] Imx. Property [1.sup.0] and Lemma 2.3 imply that End(Imx) [congruent to] End([C.sub.4]). In view of Lemma 2.8, Imx [congruent to] [C.sub.4] and there exists c [member of] G such that

G = M [lambda] <c>, Imx = <c> [congruent to] [C.sub.4], M = Kerx.

By Lemma 2.4, each y [member of] H (x) is induced by a homomorphism y : Im x [right arrow] Ker x = M. Property [2.sup.0] implies that each such homomorphism is zero, i.e., M is a 2 -group. Hence Imx = <c> is a Sylow 2-subgroup of G. In view of Lemmas 2.1 and 2.2, [absolute value of ([x])] is equal to the number of Sylow 2-subgroups of G. Since all Sylow 2-subgroups of G are conjugate, we have [absolute value of ([x])] = [G : [C.sub.G](c)] = [M : [C.sub.M](c)]. By property [5.sup.0], [M : [C.sub.M](c)] = 9. If g [member of] [C.sub.M](c), then [??] [member of] D(x). Since D(x) is a 2-group (property [4.sup.0]) and M is a 2'- group, we have [??] = 1, i. e., g [member of] Z(G) and

[C.sub.M] (c) [subset] Z(G), [C.sub.M] (c) [??] G, [M : [C.sub.M] (c)] = [absolute value of (M/[C.sub.M](c))] = 9. (7.10)

It follows that all {2, 3}-elements of G belong into the centre of G, and, therefore, G splits into a direct product

G = [G.sub.1] x [G.sub.2],

where [G.sub.1] and [G.sub.2] area {2, 3}'-subgroup and a {2, 3}-subgroup of G, respectively. Denote by z the projection of G onto its subgroup [G.sub.1]. Then z [member of] J(x) and property [3.sup.0] implies z = 0. Therefore, [G.sub.1] = <1>, G is a {2, 3}-group and M is a 3-group.

Since Im x is Abelian, [??] [member of] V(x) for each g [member of] G. Therefore, by (7.10) and properties 40, 80, we get

M/[C.sub.M](c) [congruent to] [C.sub.3] x [C.sub.3]. (7.11)

Our next aim is to prove that

[g.sup.3] = 1 for each g [member of] M. (7.12)

Assume that k is the smallest positive integer such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for each g [member of] M. To prove (7.12), it is necessary to show that k = 1. To obtain a contradiction, suppose that k > 1. Define the map y : G = M [lambda] <c> [right arrow] G as follows:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

To prove that y is an endomorphism of G choose [c.sup.i]g, [c.sup.j]h [member of] G (i, j [member of] [Z.sub.4]; g, h [member of] M). Using Lemma 2.13 and (7.11), we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Hence y is an endomorphism of G. Let us find y3 :

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

i.e., [y.sup.3] = 1 and y is an automorphism of G. Clearly, y [member of] D(x). Property 40 implies that y = 1 and, therefore, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for each g [member of] M. This contradicts the choice of k. It follows that the assumption k > 1 is false and hence k = 1, i.e., (7.12) holds.

Choose g [member of] M\[C.sub.M](c). Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. By Lemma 2.3, the map [y.sub.2] defined by

(Ker x)[y.sub.2] = <1>, c[y.sub.2] = ([g.sup.-1] cg)[y.sub.2] = [g.sup.-1][c.sup.2]g

belongs to K (x[??]). Clearly, [y.sub.2] [not equal to] 0. If [c.sup.2] [member of] Z(G), then c[y.sub.2] = [c.sup.2] and [y.sub.2] [member of] K (x) [intersection] K (x[??]). This contradicts property [6.sup.0]. Therefore, [c.sup.2] [not member of] Z(G), [[??].sup.2] [not equal to] 1 and [??] is an automorphism of order 4 of G. It follows from (7.10), (7.11), and Lemma 2.7 that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

We have also that Z(G) = [C.sub.M](c). By properties [4.sup.0] and [8.sup.0], [absolute value of (V(x))] = 72. Hence

[V(x) : [??]]= 2. (7.13)

In view of property [7.sup.0], V(x) has a subgroup B isomorphic to [G.sub.14]. We have

[V(x) : B]= 2 (7.14)

because [absolute value of (V(x))] = 72 and [absolute value of ([G.sub.14])] = 36. Since D(x) is cyclic, property [8.sup.0] implies that V(x) has only one subgroup of index 2. Therefore, by (7.13) and (7.14),

[??][congruent to] [G.sub.14] [congruent to] G/[C.sub.M](c). (7.15)

It follows from (7.1) and (7.15) that there exist a, b [member of] G such that

M/[C.sub.M] (c) = <[bar.a]> x <[bar.b]> [congruent to] [C.sub.3] x [C.sub.3],

where

[bar.a] = a x [C.sub.M](c), [bar.b] = b x [C.sub.M](c)

and

[[bar.c].sup.-1][bar.ac] = [bar.b], [[bar.c].sup.-1] [bar.b] [bar.c] = [[bar.a].sup.-1], [bar.c] = c x [C.sub.M](c).

We can choose

b = [c.sup.-1]ac.

Then

[c.sup.-1]bc = [a.sup.-1] h for some h [member of] [C.sub.M] (c).

Denote d = [h.sup.2]. Then d [member of] [C.sub.M](c) [subset] Z(G) and, by (7.12), [d.sup.2] = h. We have

[c.sup.-1] bc = [a.sup.-1] h = [a.sup.-1] [d.sup.2], [c.sup.-1] x b[d.sup.-1] x c = [a.sup.-1] d = [(a[d.sup.-1]).sup.-1], [c.sup.-1] x a[d.sup.-1] x c = [c.sup.-1] ac x [d.sup.-1] = b[d.sup.-1].

Denote the elements a[d.sup.-1] and b[d.sup.-1] by a and b, respectively. Then

[c.sup.-1] ac = b, [c.sup.-1] bc = [a.sup.-1], [a, b] [member of] [C.sub.M](c) [subset] Z(G).

Consider the subgroup

N = <c, a, b, [a, b]> = <a, b, [a, b]> [lambda] <c>

of G. Clearly, N [??] G. The factor-group G/N = ([C.sub.M](c) x N)/N is an elementary Abelian 3-group, because [g.sup.3] = 1 for each g [member of] M. Assume that N [not equal to] G. There exist h [member of] [C.sub.M] (c) and L [??] G such that N [subset] L and G/L = (hL) [congruent to] [C.sub.3]. Consider the endomorphism y of G given as follows:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII],

where [epsilon] is the natural homomorphism. By the construction of y, y [not equal to] 0 and y [member of] J(x). This contradicts property [3.sup.0]. Therefore, G = N. Since [g.sup.3] = 1 for each g [member of] M, we have

G = <c, a, b, [a, b]> = <a, b, [a, b]> [lambda] <c>,

where

[c.sup.4] = 1, [a.sup.3] = [b.sup.3] = [[a, b].sup.3] = 1, [c.sup.-1] ac = b, [c.sup.-1] bc = [a.sup.-1], [a, b] [member of] Z(G).

If [a, b] = 1, then

G = <a, b, c | [c.sup.4] = [a.sup.3] = [b.sup.3] = 1, ab = ba, [c.sup.-1] ac = b, [c.sup.-1] bc = [a.sup.-1]> = [G.sub.14].

If [a, b] [not equal to] 1, then denoting d = [a, b], we have

G = <a, b, c, d | [c.sup.4] = [a.sup.3] = [b.sup.3] = [d.sup.3] = 1, ab = bad, [c.sup.-1] ac = b, [c.sup.-1] bc = [a.sup.-1], cd = dc, ad = da, bd = db> = [G.sub.0].

The lemma is proved.

Theorem 7.1. The endomorphism semigroup of a group G is isomorphic to the endomorphism semigroup of the group [G.sub.14] if and only if G = [G.sub.14] or G = [G.sub.0]. Proof. Let G be a group such that

End(G) [congruent to] End([G.sub.14]). (7.16)

Since [G.sub.14] is finite, G is finite, too ([1], Theorem 2). By Lemma 7.3 and isomorphism (7.16), there exists x [member of] I(G) that satisfy properties [1.sup.0]-[8.sup.0] of Lemma 7.3. Lemma 7.4 implies that G is isomorphic to [G.sub.14] or [G.sub.0]. It follows from here and Lemma 7.2 that the statement of the theorem is true. The theorem is proved and so is part (4) of Theorem 1.1.

Theorem 1.1 is proved.

doi: 10.3176/proc.2016.3.06

ACKNOWLEDGEMENTS

The publication costs of this article were covered by the Estonian Academy of Sciences.

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Alar Leibak and Peeter Puusemp *

Department of Mathematics, Tallinn University of Technology, Ehitajate tee 5, 19086 Tallinn, Estonia

Received 30 June 2015, accepted 20 October 2015, available online 30 May 2016

* Corresponding author, peeter.puusemp@ttu.ee

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Title Annotation: | MATHEMATICS |
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Author: | Leibak, Alar; Puusemp, Peeter |

Publication: | Proceedings of the Estonian Academy of Sciences |

Article Type: | Report |

Date: | Sep 1, 2016 |

Words: | 7795 |

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