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Show that the intersection of normal subgroups is normal.

Let $H_1$ and $H_2$ be normal in $G$, meaning $\forall a \in G$, $aH_1 = H_1a$ and $aH_2 = H_2a$. We show that $a (H_1 \cap H_2) = (H_1 \cap H_2)a$.

Now notice $aH_1 \cap aH_2 = H_1a \cap H_2 a$, so we are done.

My book does something completely off and they showed that $(H_1 \cap H_2)$ is actually a subgroup first, then they used conjugation to prove normality. I thought showing left and right cosets coinciding is enough?

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  • $\begingroup$ A group is normal IFF it's a union of conjugacy classes. From here it's quite straightforward.. $\endgroup$ – Curious Droid Oct 25 '14 at 0:05
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You are correct, showing left and right cosets are identical is sufficient. $gH = Hg$ for all $g\in G$ is equivalent to $gHg^{-1} = H$ is equivalent to $gHg^{-1} \subseteq H$ is equivalent to $H\trianglelefteq G$. So your proof looks fine.

Clearly your book started by proving that if $H_1, H_2\le G$ then $H_1\cap H_2\le G$, something you just assumed (probably properly) in your proof. It then used the conjugation criterion rather than your choice of showing left and right cosets coincide.

So both are valid. The book chose to prove something you assumed, and chose a different but equivalent path to prove normality.

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A subgroup $H$ of $G$ is normal if and only if it is the kernel of a homomorphism.

Let $H_1$, $H_2$, ..., $H_r$ be normal subgroups of $G$, let $\pi_i$ be the canonical projection of $G$ to the quotient group $G/H_i$ and let $\pi$ be the homomorphism from $G$ to the direct product $(G/H_1)\times (G/H_2) \times \ldots\times (G/H_r)$ given by $\pi(g) = (\pi_1(g), \pi_2(g), \ldots, \pi_r(g))$.

$\pi$ is a homorphism whose kernel is the intersection of the $H_i$'s

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You want to prove $\;H_1\,,\,H_2\lhd G\implies H_1\cap H_2\lhd G\;$ , so take $\;x\in H_1\cap H_2\;,\;\;g\in G\;$ :

$$x^g:=g^{-1}xg\in\begin{cases}H_1&,\text{because}\;x\in H_1\lhd G\\{}\\H_2&,\text{because}\;x\in H_2\lhd G \end{cases}\implies g^{-1}xg\in H_1\cap H_2$$

and we're done.

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