# Normalizers and normal subgroups.

1. Apr 29, 2013

### Artusartos

If G is a group with subgroups H and K, and if H is normal, then HK=KH, right?...Since we know that N(H) = G. However, since H commutes with K, then H must also be contained in N(K), right?...Since N(K) is the set of elements that commute with K...but I was a bit confused, because we know that, for any subgroup M, GM=MG. So G must be contained in the normalizer of any subgroup, right? But that doesn't make sense, since the normalizer of any subgroup would then need to equal G and every subgroup would be normal...I know I'm probably missing something, but I'm not sure what. So I was wondering if anybody could clarify this for me.

2. Apr 29, 2013

### micromass

I don't see why $HK=KH$ would imply that $H\subseteq N(K)$.
You seem to think that from $HK=KH$ follows that for each $h\in H$ holds that $hK=Kh$. The latter seems like a much stronger statement.

The statement $HK=KH$ implies to me that for each $h\in H$ and $k\in K$, there exists a $h^\prime\in H$ and $k^\prime \in K$ such that $hk=k^\prime h^\prime$.

The statement $H\subseteq N(K)$ implies that for each $h\in H$ and $k\in K$, there exists is a $k^\prime \in K$ such that $hk=k^\prime h$. So this statement is the previous statement, but with the additional fact that $h^\prime = h$. This seems like a much stronger statement to me.

3. Apr 29, 2013

### Artusartos

Ok thanks, I didn't realize that...

4. Apr 29, 2013

### micromass

Maybe it is true that the two statements are equivalent. I just said that it looks like a stronger statement to me, but stronger statements can sometimes be proven equivalent. To conclusively end this discussion, you should look for a example where $HK=KH$, but where $H$ is not a subset of $N(K)$. Try your favorite small, nonabelian groups and maybe you'll find one.