Normal group problem

**tttcomrader** · October 9th 2007, 07:27 PM

Let N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.

I'm a confused about NH, what kind of element belong to this set?

**Opalg** · October 10th 2007, 12:34 AM

Originally Posted by tttcomrader

Let N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.

I'm a confused about NH, what kind of element belong to this set?

NH means the set of all products of an element of N times an element of H. So an element of NH is of the form nh, where n is in N and h is in H.

**ThePerfectHacker** · October 10th 2007, 10:41 AM

Originally Posted by tttcomrader

Let N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.

I'm a confused about NH, what kind of element belong to this set?

The set $NH = \{nh| n\in N \mbox{ and }h\in H\}$ . If $N,H$ are subgroups of $G$ then not necessarily does it means $NH$ is a subgroup. If however, $N$ is normal, i.e. $gng^{-1} \in N$ for any $n\in N$ and $g\in G$ then $NH$ is a group.

However, it turns out that $N$ need not be normal in $G$ . It can have a weaker property but yet $NH$ .

Theorem: If $N$ has the property that $hnh^{-1} \in N$ for any $n\in N$ and $h\in H$ then $NH$ is a group.
(Do you see why this would prove you case? Meaning, if $N$ is normal then certainly $hnh^{-1}\in N$ . So of course it works).

Proof: All I will show is closure of $NH$ which is the only part which takes a little work. Let $x_1 \in NH$ and $x_2\in NH$ we want to show $x_1x_2\in NH$ . But this means $x_1 = n_1h_1$ and $x_2 = n_2h_2$ for $n_i \in N$ and $h_i \in H$ . Thus, $x_1x_2=n_1h_1n_2h_2$ . But is this product in $NH$ ? Not necessarily, so we need to use the additional condition in the theorem. Which says $h_1n_2h_1^{-1}\in N$ , i.e. $h_1n_2h_1^{-1} = n_3\implies h_1n_2 = n_3h_1$ for some $n_3\in N$ . Thus, $n_1h_1n_2h_2 = n_1(h_1n_2)h_2 = n_1(n_3h_1)h_2 = (n_1n_3)(h_1h_2)$ which certainly is in $NH$ . Q.E.D.

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