I’m trying to disprove the following statement: “let $G$ be a group, and suppose $H,K$ are subgroups of $G$; if $H$ is normal in $G$ and $K$ is normal in $H$, then $K$ is normal in $G$.”

My counterexample is with $G=D_4$, $H=\{e,a^2,b,a^2b\}$, and $K=\{e,b\}$.

I know that $H$ is normal in $G$ because $[G:H]=2$ and I know that $K$ is normal in $H$ because $[H:K]=2$. But we have that $[G:K]=4$, with $b \notin Z(G)$. So, $K$ is not normal in $G$. Therefore, normality of subgroups is not transitive. Is this correct?

  • 3
    $\begingroup$ That's right, and this is one of the standard examples. $\endgroup$ Aug 13, 2020 at 6:07
  • 1
    $\begingroup$ Another one is given in this post. There are several more posts about it, e.g., see here. $\endgroup$ Aug 13, 2020 at 10:53


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