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