Given a finite group $G$ and its subgroups $H,K$ such that $$G \times H \cong G \times K$$ does it imply that $H=K$.

Clearly, one can see that this doesn't work out for all subgroups. Is there any condition by which this can remain true.

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    @Chandru1: Now that you've noticed the meta question, I've deleted the previous comment. It's not inconvenience at all, and please don't be offended! Your questions are interesting, and you should continue. It's just better to give some sort of "motivation" or context to these questions (like where you found the question, why you want to know, and what you have tried), so that those who answer know that it's not homework, and feel they're helping someone who is genuinely interested. Please see the advice on the meta question. – ShreevatsaR Aug 11 '10 at 23:06
up vote 10 down vote accepted

Yes. This is an easy consequence of the Krull-Schmidt theorem:

Edit: To be clear, what I am claiming is that if one has finite groups $H,K,G$ such that $G \times H \cong G \times K$, then $H \cong K$. As Steve D points out below, the hypothesis and conclusion of the OP's literal question are a bit different than this. But because of the Krull-Schmidt theorem, the OP's literal question becomes: let $H$ and $K$ be isomorphic subgroups of a finite group $G$. When do we have $H = K$? It is quite clear that the answer is "not always", and I find implausible that there would be a clean necessary and sufficient condition for this. But let's see what transpires...

  • How did you conclude this? Supposedly $H$ and $K$ are subgroups of $G$, and we are looking to show not that they are isomorphic, but equal. So the conclusion is not true: $G=C_2\times C_2$, $H$ and $K$ two different copies of $C_2$. – user641 Aug 11 '10 at 22:21
  • @Steve D: you're right, I interpreted the question to be something other than what the OP literally asked. The OP should clarify whether my interpretation is correct. – Pete L. Clark Aug 11 '10 at 22:33
  • I'm sure your interpretation is correct. But the OP said $H=K$, and then claimed he knew it didn't work for all subgroups. I am sure it is not you who is confused. – user641 Aug 11 '10 at 22:40
  • The link is broken (and I think PlanetMath has been having problems). Wikipedia has a suitable article, but I leave it to you to decide what to do. – user43208 Feb 9 '17 at 16:06

Here's an entry point into the literature, from the introduction to Lam's paper [1]:

In the study of any algebraic system in which there is a notion of a direct sum, the theme of cancellation arises very naturally: if $A \oplus B \cong A\oplus C$ in the given system, can we conclude that $B \cong C$? (For an early treatment of this problem, see the work of Jonsson and Tarski [JT] in 1947.) The answer is, perhaps not surprisingly, sometimes "yes" and sometimes "no": it all depends on the algebraic system, and it depends heavily on the choice of A as well.

Starting with a simple example, we all know that, by the Fundamental Theorem of Abelian Groups, the category of finitely generated abelian groups satisfies cancellation. But a little more is true, which solved what would have been the "Third Test Problem" for §6 in Kaplansky's book [Ka 1] (see the Notes in [Ka_1:§20]): if A is a f.g. (finitely generated) abelian group, then for any abelian groups $B$ and $C$, $A\oplus B \cong A\oplus C$ still implies $B \cong C$. Thus, f.g. abelian groups A remain "cancellable" (with respect to direct sums) in the category of all abelian groups. This takes a proof, which was first given, independently, by P. M.Cohn [Co] and E. A. Walker [W]. And yet, there exist many torsionfree abelian groups of rank 1 (that is, nonzero subgroups of the rational numbers Q ) that are not cancellable in the category of torsionfree abelian groups of finite rank, according to B. Jonsson [Jo].

[1] T.Y. Lam. A Crash Course on Stable Range, Cancellation, Substitution, and Exchange
University of California, Berkeley, Ca 94720

  • You might also mention cancellation in finite structures of the same type. Under certain mild conditions, nth roots are unique (Lovasz) and cancellation is possible. The phrase "algebra with zero" comes to mind. Perhaps you can provide more detail? [signature removed by moderator] – G. Paseman Aug 11 '10 at 21:35

Concerning Steve D's interpretation of the question, here is a partial answer.

Consider the following property (P) of a group: for any two subgroups $H$ and $K$ of $G$, if $H \cong K$, then $H = K$.

Claim: For a finite group $G$, the following are equivalent:
(i) $G$ has property (P).
(ii) $G$ is cyclic.

Cyclic groups are characterized among finite groups by having at most one subgroup of any given order, so certainly (ii) $\implies$ (i).

Conversely, assume $G$ has property (P). Then it is a Dedekind group: all of its subgroups are normal (for otherwise it has two subgroups which are conjugate -- hence isomorphic -- but unequal).

Case 1: $G$ is abelian, of exponent $N$. Then $G$ is isomorphic to $C_N \times G'$ for some subgroup $G'$ (this is a step in the classification of finite abelian groups; it also follows easily from the theorem). If $G'$ is not cyclic, then $G'$ is not trivial hence contains an element of order $p | N$ and thus over all $G$ contains at least two subgroups of order $p$.

Case 2: $G$ is a nonabelian Dedekind group, i.e., a Hamiltonian group. As Dedekind showed, a Hamiltonian group (finite or otherwise!) must contain a subgroup isomorphic to the quaternion group $Q_8$. But $Q_8$ contains three cyclic subgroups of order $4$, namely those generated by the elements $i$, $j$ and $k$.

There are some noncyclic infinite abelian groups with property (P), e.g. $\mathbb{Q}/\mathbb{Z}$, but probably one can classify them as well: I haven't thought much about it.

  • Pete, a quicker proof of your statement. If the finite group $G$ has property (P), then if $p$ divides $G$, $G$ has a unique subgroup of order $p$. Thus, if $p$ is odd, $G$ is cyclic; if $p=2$, $G$ is cyclic or generalized quaternion. Then you can simply rule out the quaternion group as you did above. – user641 Aug 12 '10 at 1:44
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    The infinite case is not much harder. The group must be torsion, and now, using your Dedekind group argument, must be abelian. We thus can split it up into p-components. It's not hard to see such a group - if infinite - is indecomposable and divisible, so must be quasicyclic. Thus every such group with (P) is abelian torsion, and its p-components are finite cyclic or quasicyclic. – user641 Aug 12 '10 at 1:57
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    So to sum up (sorry for all the comments), if we remove the "finite" restriction from the original question, such groups are exactly those with property (P), and those are fully classified up to isomorphism above. – user641 Aug 12 '10 at 2:25
  • @Steve D: thanks for your comments. Though I didn't say so above, I believe I was laboring under the misimpression that the infinite cyclic group has property (P) whereas, as you correctly point out, it does not. Were it not for this oversight I would probably have typed out the general case! – Pete L. Clark Aug 12 '10 at 3:43

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