$H_1 ,H_2 \unlhd \, G$ with $H_1 \cap H_2 = \{1_G\}$. Prove every two elements in $H_1, H_2$ commute

This is the proof, which I mostly understand except for one bit:

You have $h_1 \in H_1$ and $h_2 \in H_2$.

We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore

$$h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\}$$

and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let's first multiply everything on the left by $h_1$

$$h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$$ $$h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}$$ $$h_2^{-1}h_1h_2 = h_1 \{1_G\}$$

Multiply both sides on the left by $h_2$ giving us

$$h_1h_2 = h_2 h_1$$

The bit I don't get is right at the beginning. Why is this correct: "We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$."

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If you look at the part in parentheses, the argument shows that is an element in $H_1$, which is due directly to the fact that $H_1$ is a normal subgroup. It's clear $h_1^{-1}$ is an element, and since $H_1$ is a subgroup, it's closed under its operations. – Clayton Dec 7 '12 at 15:51

$H_1$ is normal in $G$, so for any $g\in G$ you have $g^{-1}H_1g \subseteq H_1$. So in particular with $g = h_2$ and $h_1\in H_1$: $$h_2^{-1}h_1h_2 \in H_1.$$ Then multiplying by $h_1^{-1}\in H_1$ doesn't move us outside $H_1$, so $$h_1^{-1}(h_2^{-1}h_1h_2) \in H_1.$$ About the next part (ref comment) you have the same. There you have $H_2$ a normal subgroup of $G$. So that means with $h_1 \in H \leq G$ and $h_2 \in H_2 \Rightarrow h_2^{-1} \in H_2$ you have $$h_1^{-1}h_2^{-1}h_1 \in H_2.$$ And so $$(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2.$$
Does the $h_2^{-1}h_1h_2 \in H_1$ come from the defintion of a normal subgroup? – Kaish Dec 7 '12 at 15:56
@Kaish: Yes exactly. $H$ is normal in $G$ means exactly that $g^{-1}hg \in H$ for all $h\in H$ and $g\in G$. Note that $h_2$ is here just considered as an element of $G$ (because $H_2 \leq G$). en.wikipedia.org/wiki/Normal_subgroup – Thomas Dec 7 '12 at 15:57
So in the next line, with the similary bit, would it be: $h_1^{-1}h_2^{-1}h_1 \in h_1^{-1} H_2 h_1 = H_2$? – Kaish Dec 7 '12 at 15:59