normal subgroup proof

**Danneedshelp** · October 17th 2010, 02:51 PM

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

**Bruno J.** · October 17th 2010, 03:31 PM

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

.

**Danneedshelp** · October 17th 2010, 05:37 PM

Originally Posted by Bruno J.

.

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

**Bruno J.** · October 17th 2010, 06:06 PM

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).

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