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MILNE and it remains to determine the possible homomorphisms θ : C2 → Aut(C3 × C5 ). But such a homomorphism θ is determined by the image of the nonidentity element of C2, which must be an element of order 2. Let a, b, c generate C3, C5 , C2 . Then Aut(C3 × C5) = Aut(C3 ) × Aut(C5 ), and the only nontrivial elements of Aut C3 and Aut C5 are a → a−1 and b → b−1 . Thus there are exactly 4 homomorphisms θ, and θ(c) is one of the following elements: a→a b→b a→a b → b−1 a → a−1 b→b a → a−1 . b → b−1 The groups corresponding to these homomorphisms have centres of order 30, 3 (generated by a), 5 (generated by b), and 1 respectively, and hence are nonisomorphic.

A) A simple group is indecomposable, but an indecomposable group need not be simple: it may have a normal subgroup. For example, S3 is indecomposable but has C3 as a normal subgroup. (b) A ﬁnite abelian group is indecomposable if and only if it is cyclic of prime-power order. Of course, this is obvious from the classiﬁcation, but it is not diﬃcult to prove it directly. Let G be cyclic of order pn , and suppose that G ≈ H ×H . Then H and H must be p-groups, and they can’t both be killed by pm , m < n.

But Z(G) is commutative, and it follows from the structure theory of such groups (for example) that Z(G) will contain an element of order p. p-groups. 14. A ﬁnite p-group = 1 has centre = {1}. Proof. By assumption, (G : 1) is a power of p, and it follows that (G : CG (y)) is power of p (= p0 ) for all y in the above sum. Since every other term in the sum is divisible by p, so also is (Z(G) : 1). 15. A group of order pm has normal subgroups of order pn for all n ≤ m. Proof. We use induction on m.