A natural question in group theory is to know whether or not the following implication is true:

$G \times H \simeq G \times K \Rightarrow H \simeq K$.

Of course, such a cancellation property does not hold in general, for example

$\mathbb{Z} \times \mathbb{Z} \times ( \mathbb{Z} \times \cdots) \simeq \mathbb{Z} \times \mathbb{Z} \times \cdots \simeq \mathbb{Z} \times ( \mathbb{Z} \times \cdots)$

whereas $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ are not isomorphic. However, cancellation property turns out to hold for some classes of groups. In his article On cancellation in groups, Hirshon proves that finite groups are cancellation groups, that is to say, for every groups $G,H,K$, if $G \times H \simeq G \times K$ with $G$ finite, then $H \simeq K$. Below, we present an elementary proof of a weaker statement due to Vipul Naik:

Theorem: Let $G,H,K$ be three finite groups. If $G \times H \simeq G \times K$ then $H \times K$.

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

$\displaystyle h(L,G)= \sum\limits_{N \lhd L} i(L/N,G) \hspace{1cm} (1)$

Let $G,H,K$ be three finite groups such that $G \times H \simeq G \times K$. Then

$\begin{array}{c} h(L,G \times H)=h(L,G \times K) \\ h(L,G) \cdot h(L,H)=h(L,G) \cdot h(L,K) \\ h(L,H)=h(L,K) \end{array}$

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 $K$. From $G \times H \simeq G \times K$, we deduce that $H$ and $K$ have the same cardinality, so we conclude that $H$ and $K$ are in fact isomorphic. $\square$

Because a classification of finitely-generated abelian groups or of divisible groups is known, it is easy to verify that cancellation property holds also for such groups (the classification of divisible groups was mentionned in a previous note). However, in his book Infinite abelian groups, Kaplansky mentions that the problem is still open for the class of all abelian groups.

We conclude our note by an example where the cancellation property does not hold even if the groups are all finitely-presented.

Let $\Pi$ be the following presentation

$\langle a,b,c \mid [a,b] = [a,c] = 1, b^{11}, cbc^{-1}=b^4 \rangle$.

Clearly, we have the following decomposition:

$\Pi \simeq \langle a \mid \ \rangle \times \underset{:=H}{ \underbrace{ \langle b,c \mid b^{11}=1, cbc^{-1}=b^4 \rangle }}$.

Now, we set $x=a^2c^5$ and $y= ac^2$. Because $a$ commutes with $b$ and $c$, we deduce that $[x,y]=1$. Then,

$y^2=a^4c^4=xc^{-1} \Rightarrow c=y^{-2}x$

and

$x=a(ac^2)c^3=ayc^3 \Rightarrow a=y^5x^{-2}$.

Now, let us notice that

$c^5bc^{-5} = c^4(cbc^{-1})c^{-4} = c^4b^4c^{-4} = c^3 (cbc^{-1})^4c^{-4}= \cdots = b$,

hence $[x,b] = [a^2c^5,b]=c^5bc^{-5}b^{-1}=1$. Therefore,

$\Pi \simeq \langle b,x,y \mid [x,b]=[x,y]=[y^5,b]=1, b^{11}=1, y^{-2}by^2=b^4 \rangle$.

Using $y^{-2}by^2=b^4$, it is not difficult to notice that the relation $[y^5,b]=1$ can be written as $yby^{-1}=b^5$. Then, we deduce that

$y^2b^4y^{-2}= y(yby^{-1})^4y^{-1} = (yby^{-1})^{-2}=b$

so that the relation $y^{-2}by^2=b^4$ can be removed from the presentation. Finally,

$\Pi \simeq \langle b,x,y \mid [x,y]=[x,b]=1, b^{11}=1, yby^{-1}=b^5 \rangle$.

Therefore, we find another decomposition of $\Pi$ as a direct product:

$\Pi \simeq \langle x \mid \ \rangle \times \underset{:=K}{\underbrace{ \langle b,y \mid b^{11}=1, yby^{-1}=b^5 \rangle }}$.

So $\mathbb{Z} \times H \simeq \Pi \simeq \mathbb{Z} \times K$. To conclude, it is sufficient to prove that $H$ and $K$ are not isomorphic. Notice that they are both a semi-direct product $\mathbb{Z}_{11} \rtimes \mathbb{Z}$.

Let $\varphi : H \to K$ be a morphism. It is not difficult to show that a torsion element of $K$ has to be conjugated to $b$, so, without loss of generality, we may suppose that $\varphi(b)= b^m$ for some $0 \leq m \leq 10$. Then, $\varphi(c)= b^ry^s$ for some $r,s \in \mathbb{Z}$. Let $\pi : K \to \mathbb{Z}$ be the canonical projection sending $b$ to $0$ and $y$ to $1$. If $\varphi$ is onto, then so is $\pi \circ \varphi$: because $\pi \circ \varphi(b)=0$ and $\pi \circ \varphi(c)=s$, we deduce that $s= \pm 1$. On the other hand,

$b^{4m} = \varphi (b^4) = \varphi (cbc^{-1} ) = b^r y^{\pm 1} b^m y^{ \mp 1} b^{-r} = y^{ \pm 1} b^m y^{\mp 1}$,

hence $b^{4m}= b^{5m}$ or $b^{20m}=b^m$. Therefore, $11$ has to divide $m$ and we deduce that $\varphi(b)=b^m=1$. We conclude that $\varphi$ cannot be an isomorphism.

Finally, we proved that

$\mathbb{Z} \times ( \mathbb{Z}_{11} \rtimes_{a} \mathbb{Z}) \simeq \mathbb{Z} \times ( \mathbb{Z}_{11} \rtimes_b \mathbb{Z})$,

where $a : n \mapsto 4n$ and $b : n \mapsto 5n$, but $\mathbb{Z}_{11} \rtimes_a \mathbb{Z}$ and $\mathbb{Z}_{11} \rtimes_b \mathbb{Z}$ are not isomorphic.

In particular, we deduce that:

Corollary: $\mathbb{Z}$ is not a cancellation group.