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Let $G$ be finite $p$-group. $x,y\in G$ have property that if $xyx^{-1}y^{-1} \in N$, $N$ is normal subgroup of $G$, then also $x\in N$. Prove $x=1$.

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    Taking $N=G$ we see that this also holds for $x\neq 1$. Do we need to have this for *all* normal subgroups $N$? – 2017-02-24
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    @DietrichBurde Or possibly $N$ is assumed to be a proper normal subgroup. – 2017-02-24
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    Perhaps we need the condition that $G/N$ is not abelian? If it is abelian then the commutator $[G,G]$ is a subgroup of N. – 2017-02-24

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This seems to be a false statement or rather trivially true since there is no group $G$ (except for the trivial one) with mentioned properties. I'm not sure though. Please correct me if I'm wrong.

Assume that $G$ is nontrivial and that $N$ is proper. Now we have the property

$$[x,y]\in N\Rightarrow x\in N$$

Consider quotient group $G/N$. The property implies that in $G/N$ we have

$$[x,y]=1\Rightarrow x=1$$

But $G/N$ is a nontrivial $p$-group and it is well known that nontrivial $p$-groups have nontrivial centeres. So let $z\in Z(G/N)$, $z\neq 1$. Pick any $u\neq 1$ and since $zu=uz$ then $[z,u]=1$ even though none of them is $1$. Contradiction.

Therefore either $N$ is not proper or $G$ is trivial. But for $N=G$ with $G$ nontrivial the statement is trivially false.

So the only option we are left with is that $G$ is trivial in which case $x=1$ is trivially true.

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    The question is not about all $y$;so $[x,y]=1\rigtharrow x=1$ isn't true. $y$ is specific element. – 2017-02-24
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    @SoheilMemarian Well even if you fix $y$ all of the statements are still true. Just in $G/N$ you have to consider the element $yN$ in "$[xN, yN]=1\Rightarrow xN=1$" and then you pick $zN\neq 1$ from the center and consider $zNyN=yNzN$ gives a counterexample. Right? – 2017-02-24
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This is how I understand the statement:

Let $G$ be finite p-group, $N \trianglelefteq G$, $x \in G$. Suppose $\forall y \in G$ $x y x^{-1} y^{-1} \in N \implies x \in N$. Then $x = 1$.

Now we claim this statement is false for all nontrivial finite p-group.

Let $G$ be nontrivial finite p-group. Then $1_G \ne Z(G) \trianglelefteq G$ since nontrivial finite p-group has nontrivial center. Now take $N = Z(G), x \in N \setminus 1_G$. We have $\forall y \in G$ $x y x^{-1} y^{-1} = 1 \in N$. Hence $\forall y \in G$ $x y x^{-1} y^{-1} \in N \implies x \in N$ holds. But $x \ne 1$ by construction.

So this statement is false indeed.