1. normal subgroup proof

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

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

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

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

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

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

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 $H$ and $K$ are normal subgroups of a group $G$, then $HK\leq\\G$.

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

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

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

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

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

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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$xy^{-1}=abd^{-1}c^{-1}=(ab)(d^{-1}c^{-1})$. Since $K$ is a subgroup, $d^{-1}c^{-1}\in{K}$ whenever $c,d\in{K}$. So, $xy^{-1}\in{K}$

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

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