Normal-homomorph-containing implies strictly characteristic

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal-homomorph-containing subgroup) must also satisfy the second subgroup property (i.e., strictly characteristic subgroup)
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Statement with symbols

Suppose N is a normal-homomorph-containing subgroup of a group G: for any homomorphism \varphi:N \to G such that \varphi(N) is normal in G, we have \varphi(N) \le N.

Then, N is a strictly characteristic subgroup of G: for any surjective endomorphism \alpha of G, \alpha(N) \le N.

Facts used

  1. Normality satisfies image condition: The image of a normal subgroup under a surjective homomorphism is a normal subgroup of the image.


Given: A normal subgroup N of a group G such that whenever \varphi:N \to G is a homomorphism such that \varphi(N) is normal in G, we have \varphi(N) \le N. A surjective endomorphism \alpha of G.

To prove: \alpha(N) \le N.


  1. (Given data used: N is normal in G, \alpha is a surjective endomorphism): \alpha(N) is a normal subgroup of G: By fact (1), the image \alpha(N) is a normal subgroup of \alpha(G). By surjectivity, we have \alpha(G) = G, so \alpha(N) is normal in G.
  2. (Given data used: N is normal-homomorph-containing): \alpha(N) \le N: Let \varphi:N \to G be the restriction of \alpha to N. Then \varphi(N) = \alpha(N) by definition, and by step (1), \varphi(N) is normal in G. Since N is normal-homomorph-containing, we get \varphi(N) \le N, so \alpha(N) \le N.