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I am having difficulty proving the following commutator identity.

Let $N$ be a normal subgroup of a finite p-group $G$, $x \in G$ and $H = \langle N , x \rangle$. Then $[H,H] =[N,H]$

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    You may as well assume that $G =H$. Then any commutator in $G$ has the form $[x_1n_1,x_2n_2]$ with $n_1,n_2 \in N$ and $x_1,x_2$ are powers of $x$. Since $[x_1,x_2]=1$ it follows from the commutator identities that this lies in $[G,N]$. There is no need to assume that $G$ is a finite $p$-group. – 2017-02-12
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    @DerekHolt Thanks, but it is not clear to me why every commutator is of the form you describe. To me an element of $H$ is of the form $n_1x^{e_1}n_2x^{e_2} \cdots n_kx^{e_k}$ and I don't see how to simplify the commutator of such elements to be of your form – 2017-02-13
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    But you said that $N$ is a normal subgroup. – 2017-02-13
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    Right, I see it now. $\langle x \rangle N \langle x \rangle = \langle x \rangle N$ – 2017-02-13

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Suppose that $N$ is a normal subgroup of $G$ with $x \in G$. The inclusion $[H,H] \geq [N,H]$ is clear since $N \leq H$.

Now we have that $xNx =xN$, and so every element of $H = \langle x, N \rangle$ is of the form $x_1 n_1$ for some $x_1 \in \langle x \rangle, n_1 \in N$. Fix $x_1 n_1, x_2 n_2 \in H$.

We note the following two commuatator identities, for all $x,y,z \in G$

  1. $[x, z y] = [x, y]\cdot [x, z]^y$
  2. $[x z, y] = [x, y]^z\cdot [z, y]$

Applying these identities yields, \begin{align} [x_1 n_1, x_2 n_2] &= [x_1, x_2 n_2]^{n_1}[n_1,x_1 x_2] &&\text{by 1.} \\ &= \left([x_1,n_2] [x_1 , x_2] ^{n_2}\right)^{n_1}[n_1,x_1 x_2] &&\text{by 2.} \\ &= [x_1,n_2] ^{n_1}[n_1,x_1 x_2] &&\text{since $[x_1 , x_2] = 1$} \\ &= [x_1n_1,n_2] [n_2,n_1][n_1,x_1 x_2] &&\text{by 2.} \\ &\in [N,H] \end{align} And so $[H,H] \leq [N,H]$ as desired.