The kernel of a group homomorphism is a normal subgroup

Let \varphi : G \rightarrow H be a group homomorphism and let E \leq H be a subgroup. Prove that \varphi^*[E] (the \varphi-preimage of E) is a subgroup of G. If E \trianglelefteq H, prove that \varphi^*[E] \trianglelefteq G. Deduce that \mathsf{ker}\ \varphi \trianglelefteq G.

  1. Let x,y \in \varphi^*[E]. Then \varphi(x), \varphi(y) \in E, and since E is a subgroup, \varphi(x)\varphi(y)^{-1} = \varphi(xy^{-1}) \in E by the subgroup criterion. So xy^{-1} \in \varphi^*[E], and by the subgroup criterion we have \varphi^*[E] \leq G.
  2. Suppose E is normal in H. Now let g \in G and x \in \varphi^*[E]; we have \varphi(gxg^{-1}) = \varphi(g)\varphi(x)\varphi(g)^{-1} \in \varphi(g) E \varphi(g)^{-1} = E, since E is normal. Thus gxg^{-1} \in \varphi^*[E], hence \varphi^*[E] is normal in G.
  3. 1 \leq G is trivially normal for all G, and by definition \mathsf{ker}\ \varphi = \varphi^*[1]. So \mathsf{ker}\ \varphi is normal in G.
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  • neal  On April 18, 2011 at 8:10 pm

    I think you meant phi(g) (e) phi(g)^-1 is in E

    in the 1st line of the 2nd paragraph of the proof.

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