If $H$ is any subgroup of $G$ and $N = bigcap_{ain G}a^{-1}Ha$, prove that $N triangleleft G$.

A bit surprised to see all the lengthy answers here. I would certainly be discouraged if I needed to read one of those!

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Note that $N$ is an intersection of subgroups, so $N$ is a subgroup itself.

If $g in G$, then $$g^{-1}N g = g^{-1}left(bigcap_{a in G}a^{-1} H aright) g = bigcap_{a in G} (ag)^{-1} H (ag) = N$$ so $N$ is a normal subgroup of $G$.

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There are some details to be filled in in the above calculation:

Second equality: Define $phi_g: G to G: h mapsto g^{-1}hg$. This is a bijection, so it has the property $phi_g(bigcap mathcal{C}) = bigcap phi_g(mathcal{C})$.

Third equality: Use that $G ni a mapsto ag$ is a bijection, so ${ag: a in G}=G$.

You need, first, to show that $Nle G$.

I will use the two-step subgroup test.

Fix $Hle G$.

Since $ein H$ because $Hle G$ and, for any $ain G$, we have $a^{-1}ea=e$, we can conclude that

$$einbigcap_{ain G}a^{-1}Ha=N,$$

so that $Nneqvarnothing$.

Since for all $ain G$,

$$a^{-1}Ha={a^{-1}hain Gmid hin H},$$

we have that $Nsubseteq G$.

Let $x,yin N$. Then $x,yin a^{-1}Ha$ for all $ain G$. So, fixing $ain G$, we get $x=a^{-1}h_{a,x}a$ and $y=a^{-1}h_{a,y}a$ for some $h_{a,x},h_{a,y}in H$.

Then

$$begin{align} x^{-1}&=(a^{-1}h_{a,x}a)^{-1}\ &=a^{-1}h_{a,x}^{-1}(a^{-1})^{-1}\ &= a^{-1}h_{a,x}^{-1}a\ &in a^{-1}Ha end{align}$$

as $h_{a,x}^{-1}in H$, as $Hle G$. Hence $x^{-1}in N$ since $a$ is arbitrary.

Consider

$$begin{align} xy&=(a^{-1}h_{a,x}a)(a^{-1}h_{a,y}a)\ &=(a^{-1}h_{a,x})(aa^{-1})(h_{a,y}a)\ &=a^{-1}(h_{a,x}h_{a,y})a, end{align}$$

which is in $a^{-1}Ha$ since $Hle G$ implies $h_{a,x}h_{a,y}in H$. Hence $xyin N$ as $a$ is arbitrary.

Hence $Nle G$.

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For normality, it suffices to show that $$z^{-1}Nzsubseteq N$$ for all $zin G$. To this end, fix $zin G$. Then

$$begin{align} sin z^{-1}Nz &implies sin z^{-1}left(bigcap_{ain G}a^{-1}Haright)z\ &implies sinbigcap_{ain G}z^{-1}(a^{-1}Ha)z\ &implies sinbigcap_{ain G}(az)^{-1}H(az)\ &implies sinbigcap_{bin G}b^{-1}Hbtag{1}\ &implies sin N, end{align}$$

where $(1)$ holds since right multiplication by a group element on a group is bijective. Hence $z^{-1}Nzsubseteq N$.

Hence $Nunlhd G$.

Recall that $$ bigcap_{x in J} A_j = {x | forall j in J, x in A_j }. $$ We want to show that $N triangleleft G$ ie. $$ forall g in G, gN = Ng $$ and $N$ is a subgroup of $G$ (we see that it's an intersection of subgroups of $G$ so it's a subgroup of $G$). Let $g in G, n in N$. So $$ forall a' in G, n in a'^{-1}Ha'. $$ Take $a'=ag$. We get $$ forall a in G, n in g^{-1}a^{-1}Hag $$ So $$forall a in G, gng^{-1} in a^{-1}Ha. $$ This shows that $$forall g in G,gNg^{-1} subset N.$$ Thus $$forall g in G, gN subset Ng.$$ Using $g^{-1}$ we get $$ forall g in G, g^{-1}N subset Ng^{-1} $$ so $$forall g in G, Ng subset gN.$$ Thus $$ forall g in G, gN=Ng. $$