1. Normal subgroup

Let G be a group, and let $\displaystyle H \leq G$

(That means H is the normal subgroup of G, right?)

a)Show that $\displaystyle aHa^{-1}$ is also a subgroup of G.

My proof:

Suppose that x,y is in G, so that $\displaystyle xHx^{-1}, yHy^{-1} \in aHa^{-1}$

Since H is normal in G, pick $\displaystyle Hx^{-1}=x_{1}H, yH=Hy_{1},x_{1},y_{1} \in G$

$\displaystyle (xHx^{-1})(yHy^{-1}) = x(x_{1}H)(Hy_{1}y^{-1}) = xx_{1}Hy_{1}y^{-1}$

Now, $\displaystyle (xHx^{-1})^{-1}=x^{-1}Hx \in aHa^{-1}$

Is this right?

b) If H is finite of order k, then $\displaystyle aHa^{-1}$ is also finite of order k.

2. Do you know this theorem?
A set H in a group G is a subgroup iff $\displaystyle a \in H\;\& \;b \in H \Rightarrow \quad ab^{ - 1} \in H$?

Suppose that $\displaystyle x \in aHa^{ - 1} \;\& \;y \in aHa^{ - 1} \Rightarrow \left( {\exists c \in H} \right)\left( {\exists d \in H} \right)\left[ {x = aca^{ - 1} \wedge y = ada^{ - 1} } \right]$.

What can be said about $\displaystyle xy^{ - 1}$?

b) If H is finite of order k, then $\displaystyle aHa^{-1}$ is also finite of order k.
$\displaystyle \{ a_1, ... , a_n \}$
Let the above be the elements of $\displaystyle H$.
Then,
$\displaystyle \{aa_1a^{-1},...,aa_na^{-1} \}$
Are all distint.
Because otherwise if,
$\displaystyle aa_ia^{-1} = aa_ja^{-1}$
This implies,
$\displaystyle a_i = a_j$.
Which is an impossibility for distinct elements.
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Here is a useful theorem to know.

Theorem: Let $\displaystyle H$ be a subgroup of $\displaystyle G$ the following are equivalent.
1)For all $\displaystyle g\in G$ and $\displaystyle h\in H$ implies $\displaystyle ghg^{-1} \in H$.
2)For all $\displaystyle g\in G$ we have $\displaystyle gHg^{-1}$.
3)The left and right cosets coincide, i.e. $\displaystyle aH = Ha$.

All these 3 are equivalent formulations for "normal subgroup". My favorite is #1.