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Show that the intersection of two normal subgroup of $G$ is normal subgroup of $G$.

Attempt:

Let $H,\ K \triangleleft G$.

Then, by defintion: $\forall x \in G: xH=Hx$ and $xK =Kx$

Clearly,$ e \in H$ and $e \in K$. Hence, $e \in H \cap K$. Then, $x e x^{-1} =e \in H \cap K$. Indeed, $H\cap K = \varnothing$

Recall: $\forall H,\ K \leq G: H\cap K \leq$

Then, by the one-step subgroup test:

$\forall a,\ b \in H\cap K: ab^{-1} \in H\cap K$.

Then, $x ab^{-1}x^{-1}$. Here is the part I would be really careful and indeed would like some verification as to my proof henceforth.

$xab^{-1}x^{-1}=xax^{-1}b^{-1'}=xx^{-1}a^{'}b^{-1}=ea^{'}b^{-1'}=a^{'}b^{-1'}$

And so the intersection is a normal subgroup of G.

Any help is appreciated.

Thanks in advance.

Community
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Mathematicing
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    After recognizing $H\cap K$ as a group it is enough to prove that $gxg^{-1}\in H\cap K$ for every $g\in G$ and $x\in H\cap K$. This can be done by using the fact that $H,K$ are both normal. There is no need there to look at what you call the one-step subgroup test. – drhab Apr 24 '17 at 09:47
  • @drhab Would you then conclude that my proof is valid but superfluous? – Mathematicing Apr 24 '17 at 09:52
  • I can understand the reasoning behind it, but it is very ugly. Also the notation you are using (accents) needs clarification. – drhab Apr 24 '17 at 10:16

1 Answers1

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First show that $H\cap K$ is a subgroup of $G$ using the fact that $H$ and $K$ are subgroups of $G$ (forget for now that $H, K$ are normal subgroups of $G$). For this you have to prove that $ab^{-1}\in H\cap K$ for every $a, b\in H\cap K$.

Now you have to prove that $H\cap K$ is a normal subgroup. Here use the fact that $H$ and $K$ are normal subgroups of $G$. So you can show that if $x\in G$ and $y\in H\cap K$ then $xyx^{-1}\in H\cap K$.

Prajwal Kansakar
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