proof by contradiction

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.

Quote:

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Originally Posted by deniselim17

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 $x \in A \backslash \langle h \rangle$, $x \notin \langle h \rangle N$.

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I am a tad confused with what you are trying to show here. I cannot see where the k fits in...

I interpreted it as: if x\in A\<h> then x \not\equiv h^n mod N for all n (x lies in a different coset of A/N form any power of h).

However, this is not true. Take A=C_2xC_2, and let N=<(2, 2)>. Take H=<(2, 0)> and let x=(0, 2). Then (0, 2)=(2, 0) mod N...so I think I have interpreted this incorrectly...but I can't see any other way of interpreting it...

For your second question, and abelian group will do (as then x^{-1}h^{-1}xh=1). The example I gave above works here.