1. ## normal subgroup proof

Q: Prove that if $\displaystyle H$ and $\displaystyle K$ are normal subgroups of a group $\displaystyle G$, then $\displaystyle HK\leq\\G$.

A: Since $\displaystyle H\leq\\G$ and $\displaystyle k\leq\\G$, $\displaystyle H$ and $\displaystyle K$ have the same identity element $\displaystyle e$. Thus, $\displaystyle ee=e\in{HK}$ which shows $\displaystyle HK$ is not empty.

Now, let $\displaystyle x,y\in{HK}$. Then $\displaystyle x=ab, y=cd$ and $\displaystyle y^{-1}=(cd)^{-1}=d^{-1}c^{-1}$. So,

$\displaystyle xy^{-1}=(ab)(cd)^{-1}=(ab)c^{-1}d^{-1}=a(bc^{-1})d^{-1}$.

If we let $\displaystyle x=y$, then $\displaystyle xy^{-1}=e\in{HK}$. So, HK is a subgroup of H.

Now, to show it is a normal subgroup, do I just compute $\displaystyle xHK$ and show it is equal to $\displaystyle HKx$?

I am not sure if I am showing enough in the proof I wrote down. Some help would be great.

Thanks

2. Originally Posted by Danneedshelp
Q: Prove that if $\displaystyle H$ and $\displaystyle K$ are normal subgroups of a group $\displaystyle G$, then $\displaystyle HK\leq\\G$.

A: Since $\displaystyle H\leq\\G$ and $\displaystyle k\leq\\G$, $\displaystyle H$ and $\displaystyle K$ have the same identity element $\displaystyle e$. Thus, $\displaystyle ee=e\in{HK}$ which shows $\displaystyle HK$ is not empty.

Now, let $\displaystyle x,y\in{HK}$. Then $\displaystyle x=ab, y=cd$ and $\displaystyle y^{-1}=(cd)^{-1}=d^{-1}c^{-1}$. So,

$\displaystyle xy^{-1}=(ab)(cd)^{-1}=(ab)c^{-1}d^{-1}=a(bc^{-1})d^{-1}$. Mistake here. Read your line above! In any case, what is the conclusion of this calculation? You want to show that $\displaystyle xy^{-1} \in HK$.

If we let $\displaystyle x=y$, then $\displaystyle xy^{-1}=e\in{HK}$. So, HK is a subgroup of H.

Now, to show it is a normal subgroup, do I just compute $\displaystyle xHK$ and show it is equal to $\displaystyle HKx$? Yes! You can just use the fact that $\displaystyle xHKx^{-1}=(xHx^{-1})(xKx^{-1})$...

I am not sure if I am showing enough in the proof I wrote down. Some help would be great.

Thanks
.

3. Originally Posted by Bruno J.
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$\displaystyle xy^{-1}=abd^{-1}c^{-1}=(ab)(d^{-1}c^{-1})$. Since $\displaystyle K$ is a subgroup, $\displaystyle d^{-1}c^{-1}\in{K}$ whenever $\displaystyle c,d\in{K}$. So, $\displaystyle xy^{-1}\in{K}$

4. Originally Posted by Danneedshelp
$\displaystyle xy^{-1}=abd^{-1}c^{-1}=(ab)(d^{-1}c^{-1})$. Since $\displaystyle K$ is a subgroup, $\displaystyle d^{-1}c^{-1}\in{K}$ whenever $\displaystyle c,d\in{K}$. So, $\displaystyle xy^{-1}\in{K}$
But didn't you write $\displaystyle y=cd$ so that $\displaystyle c \in H, d \in K$? Same as $\displaystyle x=ab$ with $\displaystyle a \in H, b\in K$...

You'll certainly need to use the fact that the two subgroups in question are normal. It's not true in general that $\displaystyle HK$ is a subgroup (though it's true if either is normal).