Cancellation of Direct Products

Alessandro Avellino
a.avellino2@studenti.unipi.it

Introduction

In mathematics, usually, very simple statements never have equally simple answers. Sometimes it can even happen that what seems intuitive is wrong. An example of this is that if a group is the product of a normal subgroup and its quotient, it is not always possible to find the copy of the quotient in the group itself, for example in \( Q_8 \). Though there is some theorem which consents us to assert it (Schur-Zassenhauss)

The same happens for the following statement:

Let \( S, N, T \) be 3 groups. \[ S \times T \cong N \times T \Rightarrow S \cong T. \]

This statement is false in general. However, if we assume that the 3 groups are finite, the theorem becomes surprisingly true. From now on, by groups I will implicitly mean finite groups.

Theorem:
Let \(Y_1, Y_2, Z_1, Z_2\) be 4 groups such that:

\( Z_1 \cong Z_2 \)
\(Y_1 \times Z_1 \cong Y_2 \times Z_2\)

Then \( Y_1 \cong Y_2 \).

Lemma 1:
Let \(X, A, B\) be groups. Let us denote by \(h(A,B)\) the number of homomorphisms from \(A\) to \(B\).

\[ h(X, A\times B) = h(X,A)\cdot h(X,B) \]

Indeed, it is easy to verify that there exists an injective function from \(h(X,A\times B)\rightarrow h(X,A)\times h(X,B)\) and an injective one in the other direction. Since they are finite, there is a bijection.

Lemma 2:
Let us denote by \(h'(B,A)\) the injective homomorphisms from \(B\) to \(A\).
Then for all groups \(A,X\) we have:

\[ h(X,A)=\sum_{N\lhd X} h'(X/N, A) \]

This is due to the fact that, by the first isomorphism theorem, for every homomorphism there exists and is unique an injective homomorphism from the quotient by the kernel.

Proof

Given the isomorphism we have that, for every finite group \( X \),

\[ h(X,Y_1)\cdot h(X,Z_1)=h(X, Y_1\times Z_1) = h(X, Y_2\times Z_2)=h(X, Y_2)\cdot h(X,Z_2). \]

The first equality derives from Lemma 1. The second from the isomorphism of the direct products.

We have then, from the isomorphism between \( Z_2 \) and \( Z_1 \), that \( h(X,Z_1)=h(X, Z_2) \).

From this we obtain that \( h(X,Y_1)=h(X,Y_2) \).

Now we apply Lemma 2 to obtain that:

\[ \sum_{N \lhd X} h'(X/N, Y_1)=h(X,Y_1)=h(X,Y_2)=\sum_{N \lhd X} h'(X/N, Y_2) \]

By induction on the order (size), suppose that \( h'(H,Y_1)=h'(H,Y_2) \) with

\[ \forall \, |H|<|X| \Rightarrow h(X,Y_1)-h'(X,Y_1)=\sum_{\{e\} \neq N \lhd X} h'(X/N,Y_1) \] \[ =\sum_{\{e\} \neq N \lhd X} h'(X/N,Y_2)=h(X,Y_2)-h'(X,Y_2) \]

The middle equality is by induction and the two extremes are by the previous formula. But then from this we obtain, since \( h(X,Y_1)=h(X,Y_2) \),

\[ h'(X,Y_1)=h'(X,Y_2) \]

Finally we set \( X=Y_1 \), which from the formula tells us that \( 0 \neq h'(Y_1, Y_1)=h'(Y_1, Y_2) \Rightarrow Y_1 \hookrightarrow Y_2 \) and therefore, by cardinality, \( Y_1 \cong Y_2 \).