15

Let $G\times K$ be a finite group. Suppoose $G\times K\cong H\times K$. Is this sufficient to imply $G\cong H$?

Easy
  • 4,485
  • @amWhy, no! This problem actually is arose from there. Notice that I assume $G\times K$ finite. – Easy Apr 03 '13 at 06:34
  • I deleted the auto generated comment...I see the difference. Thanks for pointing that out. – amWhy Apr 03 '13 at 06:36
  • Note that this does not hold even when $G\times K (\cong H\times K)$ is finitely presented. See my answer to the original question (or see [R. Hirshon, On Cancellation in Groups, The American Mathematical Monthly, Vol. 76, No. 9 (Nov., 1969), pp. 1037-1039]). – user1729 Apr 03 '13 at 10:11
  • @Easy: Related generalization: http://math.stackexchange.com/questions/20664/ – Watson Aug 22 '16 at 13:40

2 Answers2

27

Vipul Naik found an elementary proof:

For any finite groups $L$ and $G$, let $h(L,G)$ denote the number of homomorphisms from $L$ to $G$ and $i(L,G)$ denote the number of monomorphisms from $L$ to $G$. Notice that $$h(L,G)= \sum\limits_{N \lhd L} i(L/N,G) \hspace{1cm} (1)$$

Let $G,H,K$ be three finite groups such that $G \times H \simeq G \times K$. Then \begin{gather}h(L,G \times H)=h(L,G \times K) \\ h(L,G)h(L,H)=h(L,G)h(L,K) \\ h(L,H)=h(L,K)\end{gather} for any finite group $L$, since $h(L,G) \neq 0$. Using $(1)$, it is easy to deduce that $i(L,H)=i(L,K)$ for any finite group $L$ by induction on the cardinality of $L$. Hence $$i(H,K)=i(H,H) \neq 0.$$

Therefore, there exists a monomorphism from $H$ to $K$. Since $H$ and $K$ have the same cardinality, we deduce that $H$ and $K$ are in fact isomorphic.

Mariano Suárez-Álvarez
  • 135,076
Seirios
  • 33,157
  • Any hint on the proof by induction in the middle? I do not grasp how does the sum $\sum_{N\triangleleft L}$ behave after increasing of $|L|$ by one. – Mohamed Ali Apr 06 '19 at 08:14
  • @MohamedAli The induction hypothesis implies $i(L/N,H)=i(L/N,K)$ whenever $N$ is non-trivial for cardinality reasons, and the equality between the sums allow to conclude for the equality between the last terms unaccounted for, those corresponding to trivial $N$ : $i(L,H)=i(L,K)$. – Evariste Jul 03 '21 at 12:21
11

Yes. In fact you only need to assume that $K$ is finite; that is, finite groups are cancellable. This is a theorem due to Hirshon.

Community
  • 1
Qiaochu Yuan
  • 419,620