[Math] Prove that a conjugate of a subgroup is a subgroup

abstract-algebragroup-theorysolution-verification

Let $G$ be a group and $H$ be a subgroup of $G$ and $a\in G$ fixed, then
$$H^{a}=aHa^{-1}=\{aHa^{-1} \colon h\in H\}$$
is a subgroup of $G$.

my attempt

Identity

$aha^{-1} \in H^a$

$aea^{-1} \in H^a\ \ \ \ $ Since $e ∈ H$

$aa^{-1} \in H^a$

Therefore $aa^{-1} = e ∈ H^{a}$

Closed under the operation of $G$

Let $p,q ∈ H^{a}$ and $x,y ∈ H$

$p = axa^{-1}$ q = $aya^{-1}$

as $H$ is a group

$x*y ∈ H$

$p*q = a(xy)a^{-1} ∈ H$

Inverse

let p ∈ $H^{a}$ and $x ∈ H$

$p = axa^{-1}$

$p^{-1} = ax^{-1}a^{-1}$

Then, $p^{-1} ∈ H^a$ for all $p ∈ H$

Best Answer

  1. You should write sentences and paragraphs, not just lists of equations.
  2. Why is $pq = a(xy)a^{-1}$? Explain. (This is easy, so it will be a short explanation. But it should be explained in a class at this level.)
  3. Don't let $x,y \in H$. Instead say: Let $p, q \in H^a$. Then there are $x,y \in H$ such that $p = axa^{-1}$ and $q = aya^{-1}$. Therefore...
  4. Again, for "Inverse", don't say let $p \in H^a$ and $x \in H$. Instead, say: Let $p \in H^a$. Then there is an $x \in H$ such that $p = axa^{-1}$. Therefore...
  5. Why is $p^{-1} = ax^{-1}a^{-1}$? Explain.
  6. Why is $p^{-1} \in H^a$? Explain. (You have said $p^{-1} = ax^{-1}a^{-1}$. So? Explain why this implies $p^{-1} \in H^a$. Hint: $x^{-1} \in$...)

"Identity" is a mess. You wrote: $H^a = aha^{-1}$. No. No! $H^a$ is a subset of $G$. $aha^{-1}$ is a single element of $G$. They are not the same kind of thing: one is a set, the other is an element. They are not equal. Are you trying to say that $H^a$ is the set of $aha^{-1}$, for all $h \in H$? If that's what you want to say, then say that. Although I don't know why you would want to say that. If what you want to say is something else, then say that instead.

In the next line you write $H^a = aea^{-1}$ since $e \in H$. No!! Are you trying to say that $H^a$ includes the element $aea^{-1}$? There is a correct way to say that. You seem to know what that correct way is: you wrote $e \in H$. And indeed, you can write $aea^{-1} \in H^a$.

Well, I'm sorry to be critical of how you wrote this. You clearly understand the main mathematical ideas. I hope that you will write your proof more carefully.

Related Question