Normalizers and normal subgroups.

[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.

Thanks in advance

 

[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.

 

[Artusartos]

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.

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

 

[micromass]

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

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.

 

[blue_raver22]

.

I can confirm that your understanding is correct. The normalizer of a subgroup is the set of elements in the group that commute with all elements of the subgroup. Therefore, if H is normal, it will commute with all elements of K, and thus must be contained in N(K). Similarly, since K is normal, it will commute with all elements of H, and thus must be contained in N(H). This leads to the conclusion that HK=KH, as both subgroups are contained in the normalizer of the other.

However, it is important to note that not all subgroups are normal. Only certain subgroups, such as the center of the group or the trivial subgroup, are normal. In general, the normalizer of a subgroup will not equal the entire group. So while G may be contained in the normalizer of any subgroup, it does not mean that every subgroup is normal.

I hope this clarifies any confusion you may have had. Keep questioning and exploring, as that is the essence of being a scientist!