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daveyp225
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Some confusion here reviewing some algebra.
From what I remember, a homomorphism must be specified in order to claim that a group is the semidirect product of H and K, say [itex]G \cong H \rtimes_{\phi}\,\, K[/itex], where H is normal and [itex]\phi:K \to Aut(H)[/itex] is a homomorphism.
However, I've seen solutions to problems where the subscript is omitted, such as when determining the number of groups of some order. Is it implied then? I always assume it is an inner automorphism, but I know in general that is not true. Which leads me to my next question...
Wikipedia states
But if I have H and H' are isomorphic (via isomorphism f), normal N and N' are isomorphic (via isomorphism g) with G=HN, G'=H'N'. How is G not isomorphic to G'? I would think sending x = hn -> f(h)g(n) = h'n' = x' would be an isomorphism.
From what I remember, a homomorphism must be specified in order to claim that a group is the semidirect product of H and K, say [itex]G \cong H \rtimes_{\phi}\,\, K[/itex], where H is normal and [itex]\phi:K \to Aut(H)[/itex] is a homomorphism.
However, I've seen solutions to problems where the subscript is omitted, such as when determining the number of groups of some order. Is it implied then? I always assume it is an inner automorphism, but I know in general that is not true. Which leads me to my next question...
Wikipedia states
"Note that, as opposed to the case with the direct product, a semidirect product of two groups is not, in general, unique; if G and G′ are two groups which both contain isomorphic copies of N as a normal subgroup and H as a subgroup, and both are a semidirect product of N and H, then it does not follow that G and G′ are isomorphic. This remark leads to an extension problem, of describing the possibilities."
But if I have H and H' are isomorphic (via isomorphism f), normal N and N' are isomorphic (via isomorphism g) with G=HN, G'=H'N'. How is G not isomorphic to G'? I would think sending x = hn -> f(h)g(n) = h'n' = x' would be an isomorphism.
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