# Proof of Zassenhaus Lemma

Lemma 4.52 from Advanced Modern Algebra by Rotman:

Given four subgroups $A \triangleleft A^*$ and $B \triangleleft B^*$ of a group G, then $A(A^* \cap B) \triangleleft A(A^* \cap B^*)$, $B(B^* \cap A) \triangleleft B(B^* \cap A*)$, and there is an isomorphism

$$\frac{A(A^* \cap B^*)}{A(A^* \cap B)} \cong \frac{B(B^* \cap A^*)}{B(B^* \cap A)}$$

Proof

We claim that $*(A \cap B^*) \triangleleft (A^* \cap B^*)$: that is, if $c \in A \cap B^*$ and $x \in A^* \cap B^*$, then $xcx^{-1} \in A \cap B^*$. Now $xcx^{-1} \in A$ because $c \in A$, $x \in A^*$, and $A \triangleleft A^*$; but also $xcx^{-1} \in B^*$, because $c,x \in B^*$. Hence, $(A \cap B^*) \triangleleft (A^* \cap B^*)$; similarly, $(A^* \cap B) \triangleleft (A^* \cap B^*)$. Therefore, the subset D, defined by $D = (A \cap B^*)(A^* \cap B)$, is a normal subgroup of $A^* \cap B^*$, because it is generated by two normal subgroups.

Using the symmetry in the remark, it suffices to show that there is an isomorphism

$$\frac{A(A^* \cap B^*)}{A(A^* \cap B)} \rightarrow \frac{A^* \cap B^*}{D}.$$

Define $\phi: A(A^* \cap B^*) \rightarrow (A^* \cap B^*)/D$ by $\phi: ax \rightarrow xD$, where $a \in A$ and $x \in A^* \cap B^*$. Now $\phi$ is well-defined: if ax=a'x', where $ax=a'x'$, where $a' \in A$ and $x' \in A^* \cap B^*$, then $(a')^{-1}a = x'x^{-1} \in A \cap (A^* \cap B^*) = A \cap B^* \subseteq D$. Also, $\phi$ is a homomorphism: $axa'x' = a''xx'$, where $a''=a(xa'x^{-1}) \in A$ (because $A \triangleleft A^*)$, and so $\phi(axa'x') = \phi(a''xx') = xx'D = \phi(ax)\phi(a'x').$ It is routine to check that $\phi$ is surjective and that $ker \phi = A(A^* \cap B)$. The First Isomorphism Thoerem completes the proof.

I have some problems with seeing that the kernel is $A(A^* \cap B)$. I understand that any element that belongs to $A(A^* \cap B)$ is of course mapped to the identity. But how do we know that these are the only elements in the kernel? Because we also know that $(A \cap B^*) \subseteq *A^* \cap B^*$, and any element that belongs to $A(A \cap B^*)$ must also be mapped to the identity, right?

Suppose $ax$ is in the kernel of $\phi$. This means that $x\in D=(A\cap B^*)(A^*\cap B)$. Hence, we can write $x=a'x'$ where $a'\in A\cap B^*\subset A$ and $x'\in A^*\cap B$. It follows that $ax=aa'x'\in A(A^*\cap B)$.