I have seen examples of subgroups $A \lhd B \lhd C$ such that $A$ isnt normal in $C$ (see for example this). However I tried to prove this statement and could not see the flaw in my reasoning. Here it is:

Let $A \lhd B \lhd C$. Now take $a \in A$ and $c \in C$. Then since $a$ is also contained in $B$ and $B$ is normal in $C$ it follows that $cac^{-1} \in B$.

Since $A$ is normal in $B$, we have that the quotient map $q: B \rightarrow B/A$ is a homomorphism with kernel $A$. Then we have $q(cac^{-1}) = q(c) q(a) q(c^{-1}) = q(c) q(c^{-1}) = 1$. Hence $cac^{-1}$ lies in $A$ and therefore $A$ is normal in $C$.

Thank you for your help!

  • 11
    $\begingroup$ The flaw is here: $q(c)$ is not well defined for $c \in C \setminus B$. $\endgroup$
    – Crostul
    Sep 9, 2022 at 9:49
  • $\begingroup$ ohh, yes. Thanks! $\endgroup$
    – tor
    Sep 9, 2022 at 9:51
  • 3
    $\begingroup$ So.. if you know actual specific examples of groups where the claimed result fails, then an easy way to check the argument to find the mistakes is to run it through your examples! Use explicit examples of $A$, $B$, $C$, $c$, and $a$ and see where the argument breaks down. $\endgroup$ Sep 9, 2022 at 13:45


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