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Consider the Heisenberg group $H(\mathbb Z)=\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix}$ with generators $$a = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, b = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}, c = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ Which of the following is a normal subgroup of $H(\mathbb Z)?$ $K_1=\langle a \rangle, K_2=\langle a,c \rangle, K_3=\langle c \rangle.$

My attempt: Let $h \in H(\mathbb Z), k \in K_1,$ then $$hah^{-1}=\begin{bmatrix} 1 & 1 & -y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ How to determine whether $hah^{-1}$ is in $K_1,K_2,K_3$ or not?

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  • $\begingroup$ The third one is immediately normal cause it is contained in Center fo Heisenberg group. For the first two, do as AnalysisStudent0414 wrote. $\endgroup$
    – robin3210
    Oct 7, 2019 at 15:56

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A subgroup $K$ is normal in $H=H(\mathbb{Z})$ if for all $h \in H$, $k \in K$ it holds that $k^h = hkh^{-1} \in K$.

So, for $K_1$, you got that $hah^{-1}$ is of that form. What do the elements of $K_1$ (i.e. the powers of $a$) look like? A quick computation shows that $$a^n = \begin{bmatrix} 1 & n & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

which means that, whenever $y \neq 0$, $hah^{-1} \not\in K_1$, and so $K_1$ is not normal.

You can repeat a similar argument for $K_3$, looking at the general expression for $c^n$.

$K_2$ is not cyclic, so it is a little more complicated, as you would theoretically need to check all words in $a$, $c$... but, luckily, $a$ and $c$ commute! So any element of $K_2$ can be written as $a^i c^j$ for some $i,j \in \mathbb{N}$. Now, a maybe less immediate computation shows that

$$a^i c^j = \begin{bmatrix} 1 & i & j \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

If you check whether this form is preserved if you conjugate by an arbitrary $h \in H$, this is going to tell you if $K_2$ is normal or not.

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  • $\begingroup$ Are $i,j$ integers? Because in that case if we let $i=1, j=-y$, we get that $ℎ𝑎ℎ^{−1}$ is in $𝐾_2$... $\endgroup$
    – dxdydz
    Oct 7, 2019 at 15:49
  • $\begingroup$ Precisely! So you did not find a counterexample by conjugating $a$. This is not enough to say that $K_2$ is normal, because it contains other elements $c, ac, a^2 c$ etc. You need to compute $ha^ic^j h^{-1}$ and check if it is still of the form $a^p c^q$ for some $p, q \in \mathbb{Z}$. $\endgroup$ Oct 7, 2019 at 15:51
  • $\begingroup$ I got $ha^ic^jh^{-1}=\begin{pmatrix}1&i&-iy+y+j-z\\ 0&1&-y\\ 0&0&1\end{pmatrix},$ so it's not of the form $𝑎^𝑝𝑐^𝑞$, therefore $𝐾_2$ is not a normal subgroup right? $\endgroup$
    – dxdydz
    Oct 7, 2019 at 15:57
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    $\begingroup$ Hmmmmm... Double-check your math! $\endgroup$ Oct 7, 2019 at 16:02

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