Is every subgroup of a normal subgroup normal ?
That is if $H$ is a normal subgroup of a group $G$ and $K$ is a subgroup of $H$, then $K$ is a normal subgroup of $G$. Is it true ? If not what is the example?

$a\in G$ and $k\in K$. Then $k\in H$, since $K\subseteq H$.
Now, $aka^{-1}=k_1aa^{-1}=k_1\in K$ [since $H$ is normal in $G$, $ak=k_1a$]

This implies that $K$ is normal in $H$.
Is my approach correct ?

  • 4
    $\begingroup$ The problem with you proof is that $H\lhd G$ only gives you $k_1\in H$ in you expression, not $k_1\in K$ as you claim. For that you would need to know $K\lhd G$, but that is just what you wanted to prove (and which is not true in general). $\endgroup$ May 10, 2015 at 14:35
  • 2
    $\begingroup$ This is the danger with abbreviating your reasoning. You've left out a qualification in your explanation regarding $k_1$: "Since $H$ is normal in $G$, $ak=k_1a$ for some $k_1 \in \ldots$" $\endgroup$
    – Théophile
    May 10, 2015 at 16:41

2 Answers 2


The silly counterexample is this: if $H$ is not normal in $G$, then we have $$H \not\lhd G\quad G\lhd G$$ Indeed, this need not even be true if $K$ itself is normal in $H$. For example, in $S_4$, we have $$C_2 \lhd V_4\lhd S_4$$ but $C_2\not\lhd S_4$. (Here, $V_4 = \{(1), (12)(34),(13)(24),(14)(23)\}$ and $C_2 = \{(1), (12)(34)\}$)

The flaw in your argument is taking $ak = k_1 a$ where $k_1\in K$. The fact that $a\in G$ and $H \lhd G$ only allows you to assume that $k_1 \in H$.


G is a normal subgroup of itself, but it might have subgroups that are not normal.

  • $\begingroup$ @MK please give an example $\endgroup$
    – MTMA
    May 10, 2015 at 12:03
  • 9
    $\begingroup$ Take any G with a subgroup that is not normal. $\endgroup$ May 10, 2015 at 12:09
  • $\begingroup$ Which step is wrong in my process? Please tell me. I have confusion $\endgroup$
    – MTMA
    May 10, 2015 at 12:27
  • 1
    $\begingroup$ @MTMA You are wrong because $k_1$ may be in $H$ but not in $K$ $\endgroup$
    – Rescy_
    Nov 24, 2015 at 1:21

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