6
$\begingroup$

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 $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$."

  • 0
    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.2012-12-07

1 Answers 1

6

$H_1$ 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 $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2.$

  • 0
    @Kaish: I edited.2012-12-07