Cancellation of Direct Products

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.

Solution 1:

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

http://planetmath.org/encyclopedia/KrullRemakSchmidtTheorem.html

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...

Solution 2:

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
http://math.berkeley.edu/~lam/ohio.ps

Solution 3:

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.