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?
 A: 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.
