Short problem on group theory q.3

[betty2301]

urgent due in 13 hrs]short problem on group theory q.3

[due now

 

[jbunniii]

You don't know anything about the orders of the groups, so Sylow has no relevance here.

The proof is quite elementary. I broke it down into four bite-sized claims that you can prove:

Let H = A \cap B and g = ab \in AB.

Claim 1: bHb^{-1} \subset A
Claim 2: bHb^{-1} \subset B
(Therefore bHb^{-1} \subset A \cap B)
Claim 3: abHb^{-1}a^{-1} \subset A
Claim 4: abHb^{-1}a^{-1} \subset B
(Therefore abHb^{-1}a^{-1} \subset A \cap B)

 

[betty2301]

thanks