Intersection of normal subgroups proof 
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?
 A: 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.
A: 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.
A: 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
A: Let  and  be two normal subgroups of group  
∩ <  because < and <.
Now,
Let n ∈ ∩ ⇒  ∈  and n ∈ 
and Let a ∈ 
Here,   a.n.a-¹ ∈ H, because H is a normal
subgroup
also ,  a.n.a-¹ ∈ K, because K is a normal
subgroup
⇒    a.n.a-¹ ∈ H∩K
hence H∩K is a normal subgroup.
