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I'm going through a past paper on Group theory and I wondered if someone could help with a solution to this.

Let $A$, $B$ be subgroups of $G$ such that $A\triangleleft B$ and $B/A$ is abelian. Also $N\triangleleft G$. I proved that $AN\triangleleft BN$ and now I need to show that $BN/AN$ is abelian.

Thanks

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2 Answers 2

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All cosets in $BN/AN$ are of the form $bAN$, where $b \in B$. Then for any $b, b' \in B$ we have

$bANb'AN = bAb'NAN = bAb'AN = b'AbAN = b'AbNAN = b'ANbAN$ .

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By the Isomorphism Theorems, since $A\subset B$ so $BA=B$, we have: $$\frac{BN}{AN} = \frac{B(AN)}{AN} \cong \frac{B}{B\cap AN}.$$ Since $A\leq AN$, then $A = B\cap A\leq B\cap AN$. Hence $$\frac{B}{B\cap AN}$$ is a quotient of $B/A$, hence abelian since $B/A$ is abelian.

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