Proof Verification: Abstract Algebra Groups

Question

Given that H is a group and $\ xax^{-1}∈H$ iff $a∈H$, prove that $\ x^{-1}ax∈H$ if $a∈H$.

My Attempt- If $b∈H$ then $b= x^{-1}(xbx^{-1})x∈H$. If $b∈H$, $xbx^{-1}∈H$. Setting a=$xbx^{-1}$ we get $x^{-1}ax∈H$ for all $a∈H$

EDIT- Someone pointed out that the proof was incomplete as I had to prove any element $a$ could be represented as $xbx^{-1}$. I am able to prove this. Can anyone show me how to prove this and complete the proof

I would like to verify two things in this proof. First, Is the proof logically correct? Second, Is it presented in a rigorous enough manner? The second question is because I'm new to Abstract Algebra and don't know much about the formal language used in proofs. I should also mention that this isn't the entire problem.

Answer 1

Your exercise probably states that $H$ is a subgroup of $G$ and that

for every $a\in G$, $xax^{-1}\in H$, for every $x\in G$, if and only if $a\in H$.

It could be useful to rewrite this in a more formal fashion: $$ \forall a\in G\bigl( (\forall x\in G(xax^{-1}\in H))\iff a\in H \bigr) $$

Let's examine what you do. You take $b\in H$ and observe that, for every $x\in G$, $$ b=x^{-1}(xbx^{-1})x\in H $$ Thus you can rightly conclude that $xbx^{-1}\in H$, but this is something you already know from the hypothesis. Setting $a=xbx^{-1}$ doesn't prove what you need unless you prove first that every element of $H$ can be written that way.

However your idea can be salvaged, with a slight, but decisive, change. Let $a\in H$; then, for every $x\in G$, $$ x(x^{-1}ax)x^{-1}=a\in H $$ and therefore $x^{-1}ax\in H$.

Suppose conversely that $x^{-1}ax\in H$, for every $x\in G$. Then, taking $x=1$, we get that $a\in H$.

Answer 2

If $H \le G$ and given condition says that $xax^{-1} \in H$ iff $a \in H$, then clearly, $xHx^{-1} \subset H$ for each $x \in G$ i.e.

$H$ is normal subgroup of $G$