- #1

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Thanks in advance

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- Thread starter Artusartos
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- #1

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Thanks in advance

- #2

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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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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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