MHB 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.