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 November 11th, 2014, 05:32 PM #1 Newbie   Joined: Feb 2014 Posts: 6 Thanks: 1 Help with a problem of isomorphisms. Let $H\leq G$ and are $N(H)$, $C(H)$ and $I(H)$ the normalizer of $H$, the centralizer of $H$ and the group of all inner automorphisms of $H$, respectively. Show that $N(H)/C(H)\cong I(H)$. Last edited by robinhg; November 11th, 2014 at 05:34 PM.
 January 26th, 2015, 07:59 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Let $\phi_g\in\mathrm I(G)$ be the inner automorphism induced by $g\in G$ i.e. $\phi_g(x)=gxg^{-1}$ for all $x\in G$. Now define $f:\mathrm N(H)\to\mathrm I(H)$ by $f(n)=\phi_n$ for $n\in\mathrm N(H)$ (check that this definition makes sense) and show that $f$ is a surjective homomorphism with kernel $\mathrm C(H)$.

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