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Question: $$G=\left\{\begin{bmatrix} 1&a&b \\0&1&c\\0&0&1\end{bmatrix},a,b,c\in Z_{p}\right\}$$

Prove that $G=(Z_{p}\times Z_{p})\rtimes Z_{p}$.

So for our purpose,we must find a normal subgroup of $G$ which is isomorphic to $Z_{p}\times Z_{p}$.

But I know that center of this group is $\begin{bmatrix} 1&0&b \\0&1&0\\0&0&1\end{bmatrix}$ which is isomorphic to $Z_{p}$.

Only this group doesn't satisfy our requirement. I don't know how to find this normal subgroup with order $p^{2}$.

Community
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Jack
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2 Answers2

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Let $$N:=\left\{\left[\begin{matrix}1 & a & b\\0&1&0\\0&0&1\end{matrix}\right]: a,b\in\mathbb{F}_p\right\}\simeq\mathbb{Z}_p\times\mathbb{Z}_p, $$ and let $$H:=\left\{\left[\begin{matrix}1 & 0 & 0\\0&1&c\\0&0&1\end{matrix}\right]: c\in\mathbb{F}_p\right\}\simeq\mathbb{Z}_p;$$ then $N$ is normal and $G$ is the split extension of $N$ by $H$.

ancient mathematician
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This group is the Heisenberg group over $\mathbb{F}_p$ of order $p^3$. Now every non-abelian group of order $p^3$ with a prime $p>2$ is a semidirect product. To see this, we prove it for the case that we have at least two different elements $a,b$ of order $p$. This is the case for the Heisenberg group, for which all non-trivial elements have order $p$. Then let $N=\langle a,b\rangle$ be the product of the cyclic groups $\langle a\rangle$ and $\langle b\rangle$. Let $Q=\langle c \rangle$ with another element $c$ of order $p$. Define $\theta\colon Q\rightarrow {\rm Aut}(N)$ to be the homomorphism such that $$ \theta(c^i)(a)=ab^i,\; \theta(c^i)(b)=b. $$ The group $G:=N\ltimes_{\theta}Q$ is of order $p^3$, with generators $a,b,c$ and defining relations $$ a^p=b^p=c^p=1,\; ab=cac^{-1},\; [b,a]=[b,c]=1, $$ where $[g,h]:=g^{-1}h^{-1}gh$ denotes the commutator of two elements. This group is isomorphic to the Heisenberg group over $\mathbb{Z}/(p)$. So we have $$ G\cong N\ltimes_{\theta}Q\cong (\mathbb{Z}/(p)\times \mathbb{Z}/(p)) \ltimes \mathbb{Z}/(p). $$

When $p=2$, then $G\cong D_4$, in which case the claim is also true, as we know from the dihedral group $D_4$.

Dietrich Burde
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  • beautiful answer! appreciate it. – Jack Mar 17 '17 at 15:49
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    Letting $N = \langle a,b\rangle$ for two different elements $a$ and $b$ of order $p$ is not sufficient for $N$ to have order $p^2$: a silly reason is that $b$ could be a power of $a$ other than $a$ (and $1$), and a more important reason is that $a$ and $b$ might not commute. Start by picking $a$ or $b$ as a nontrivial element in the center of $G$ (which has order $p$) and then let the other element of that pair be taken from outside the center. – KCd Apr 14 '23 at 20:20