Answer to Question #192150 in Abstract Algebra for Arvind kumar

Question

Use the fundamental theorem of homomorphism for groups to prove the following theorem, which is called the zassenhaus

Answer 1

Zassenhaus's Lemma:

Statement:

Let G be a group; let $H,H' ,K ,K'$  be subgroups of G such that H' is a normal subgroup of H and K' is a normal subgroup of K. Then $H'\cdot (H\cap K')$  is a normal subgroup of $H'\cdot(H\cap K)$ ; likewise, $K'\cdot (K\cap H')$  is a normal subgroup of $K'\cdot(H\cap K)$  ; furthermore, the quotient groups

$$(H'\cdot (H\cap K))/(H'\cdot (H\cap K'))$$

and

$$(K'\cdot(H\cap K))/(K'\cdot (K\cap H'))$$

are isomorphic.

Proof:

We first note that $H\cap K$  is a subgroup of H. Let $\eta$  be the canonical homomorphism from H to H/H'. Then $(\eta^{-1}\circ\eta)(H\cap K)=H'\cdot (H\cap K)$ , so this indeed a group. Also, note that $H\cap K'$  is a normal subgroup of $H\cap K$ . Hence

$$(\eta^{-1}\circ\eta)(H\cap K')=H'\cdot (H\cap K')$$

is a normal subgroup of

$$(\eta^{-1}\circ\eta)(H\cap K)=H'\cdot (H\cap K)$$

Now,

Let $\lambda$ be the canonical homomorphism from $H'\cdot(H\cap K)$  to $(H'\cdot(H\cap K))/(H'\cdot (H\cap K'))$ . Now, note that

$$(H\cap K )\cap(H'\cdot (H\cap K'))=(H'\cap K)\cdot (H\cap K')$$

Thus by the group homomorphism theorems, groups $(H'\cdot (H\cap K))/(H'\cdot (H\cap K'))$  and $(H\cap K)/((H'\cap K)\cdot(H\cap K'))$ are isomorphic. The lemma then follows from symmetry between H and K.