Prove One-to-One Correspondence Conjugate Subgroups & Cosets N

[Kiwi1]

Q. Prove there is a one-to-one correspondence between the set of conjugates of H and the set of cosets of N.

I have a solution to this below but am not sure if it is correct. In particular I am not sure if my definition 'f' is satisfactory. This is self study and not any kind of homework.

I have:
$$aHa^{-1}=\{axa^{-1}:x\epsilon H\}$$
$$N=\{a \epsilon G: axa^{-1} \epsilon H$$ for every $$x \epsilon H \}$$

I have proven as part of the previous question:
Q7. For any $$a,b \epsilon G, aHa^{-1}= bHb^{-1}$$ iff $$b^{-1}a \epsilon N$$ iff $$aN=bN$$

SOLUTION
define the set of conjugates: $$X=\{aHa^{-1}:a \epsilon G \}$$
define the set of cosets of N: $$Y=\{aN:a \epsilon G \}$$

define $$f:X \rightarrow Y$$ by $$f(aHa^{-1})=aN$$

f is well defined because if $$aHa^{-1}=bHb^{-1}$$ then aN = bN by Q7.

Surjectivity:
for any $$aN \epsilon Y$$
$$f(aHa^{-1})=aN$$
where $$aHa^{-1} \epsilon X$$
therefore f is surjective

Injectivity:
suppose aN = bN
$$\therefore f(aHa^{-1})=f(bHb^{-1})$$
where $$aHa^{-1}=bHb^{-1}$$ by Q7.
therefore f is injective

There is a bijection f from X to Y so there is a 1:1 correspondence between the members of X and Y.

Is this solution reasonably rigorous?

 

[caffeinemachine]

Q. Prove there is a one-to-one correspondence between the set of conjugates of H and the set of cosets of N.

I have a solution to this below but am not sure if it is correct. In particular I am not sure if my definition 'f' is satisfactory. This is self study and not any kind of homework.

I have:
$$aHa^{-1}=\{axa^{-1}:x\epsilon H\}$$
$$N=\{a \epsilon G: axa^{-1} \epsilon H$$ for every $$x \epsilon H \}$$

I have proven as part of the previous question:
Q7. For any $$a,b \epsilon G, aHa^{-1}= bHb^{-1}$$ iff $$b^{-1}a \epsilon N$$ iff $$aN=bN$$

SOLUTION
define the set of conjugates: $$X=\{aHa^{-1}:a \epsilon G \}$$
define the set of cosets of N: $$Y=\{aN:a \epsilon G \}$$

define $$f:X \rightarrow Y$$ by $$f(aHa^{-1})=aN$$

f is well defined because if $$aHa^{-1}=bHb^{-1}$$ then aN = bN by Q7.

Surjectivity:
for any $$aN \epsilon Y$$
$$f(aHa^{-1})=aN$$
where $$aHa^{-1} \epsilon X$$
therefore f is surjective

Injectivity:
suppose aN = bN
$$\therefore f(aHa^{-1})=f(bHb^{-1})$$
where $$aHa^{-1}=bHb^{-1}$$ by Q7.
therefore f is injective

There is a bijection f from X to Y so there is a 1:1 correspondence between the members of X and Y.

Is this solution reasonably rigorous?

Your solution is fine. If you know group actions, then I can tell you a shorter solution.