Normalizers and normal subgroups.

  • Thread starter Artusartos
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
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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
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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...
 
  • #4
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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.
 

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