First problem:

Let be a group with a normal subgroup of finite index .

Suppose conjugate to such that $h=a^{-1}ka$ for some .

Let $A \rightarrow \bar{A}=A/N$.

Show that for any $x \in A \backslash \langle h \rangle$, $x \notin \langle h \rangle N$.

I tried to show by contradiction.

Suppose that $x \in \langle h \rangle N$.

It is clear that $z=axa^{-1} \neq k$ since $x \notin \langle h \rangle$.

Then under the mapping, $\bar{z}=zN=\bar{a}\bar{x}\bar{a}^{-1}$.

Since $x \in \langle h \rangle N$, we can find some such that $\bar{x}=\bar{h^{m}}$.

Well, if is trivial, we have $\bar{z}=\bar{x}=\bar{h^{m}}=\bar{k^{m}}$.

If $a \notin N$, $\bar{z}=zN=axa^{-1}N=aNxNa^{-1}N=\bar{a}\bar{h^{m}}\bar{a^{-1}}=\bar{k^{m}}$.

How about when $a \in N$?

Is it true that $\bar{z}=\bar{h^{m}} \neq \bar{k^{m}}$?

Second problem:

Let be a group with a normal subgroup of finite index and $x \in A \backslash \langle h \rangle$ such that $x \in \langle h \rangle N$. Is it true that $x^{-1}h^{-1}xh \notin \langle h \rangle$?

I tried to show by contradiction. But so far I can't get any contradiction.