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I have to do two proofs.

1) If G is abelian, the the factor group G/H is abelian.

2) If H and K are normal in G, then H intersect K is normal in G.

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What is G/H is normal in? – Amr Dec 14 '12 at 17:27
I think the OP meant to say G/H is ableian. Just a slip of words. – Dan Rust Dec 14 '12 at 17:28
@Susan, you've asked 4 questions related to group theory in the last 20 minutes or so. I think it is about time you show some self effort, some ideas, insights, etc. in your own work. – DonAntonio Dec 14 '12 at 17:31
@Susan I think you should be fine with showing that G/H is abelian as long as you remember the definition of the factor group and what it means for a subset of G to be a coset. Try writing those definitions down and seeing if that helps. – Dan Rust Dec 14 '12 at 17:31
If you Google for intersection "normal subgroups", you'll get a few useful hits for the second question. – Martin Sleziak Dec 14 '12 at 17:47

1) Hint: Here you just need to know how you compose elements in $G/H$: $$ (gH)(hH) = (gh)H $$

2) You want to show that for all $g\in G$, $$g(H\cap K)g^{-1}\subseteq H\cap K.$$

So let $g\in G$. Now $gNg^{-1} \subseteq N$ and $gKg^{-1} \subseteq K$ because both $H$ and $K$ are normal in $G$. Since $$ \begin{align} H\cap K &\subseteq H \\ H\cap K &\subseteq K \end{align} $$ we have $$ \begin{align} g(H\cap K)g^{-1} &\subseteq gHg^{-1} \subseteq H\quad \text{and}\\ g(H\cap K)g^{-1} &\subseteq gKg^{-1} \subseteq K. \end{align} $$ So $g(H\cap K)g^{-1}$ is contained in both $H$ and $K$, hence $$ g(H\cap K)g^{-1} \subseteq H\cap K. $$

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