# Normal subgroup of Heisenberg group $H(\mathbb Z)$

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?

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

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.

• Are $i,j$ integers? Because in that case if we let $i=1, j=-y$, we get that $ℎ𝑎ℎ^{−1}$ is in $𝐾_2$... Oct 7, 2019 at 15:49
• 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 $p, q \in \mathbb{Z}$. Oct 7, 2019 at 15:51
• 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? Oct 7, 2019 at 15:57
• Hmmmmm... Double-check your math! Oct 7, 2019 at 16:02