# Orders of elements and homomorphisms.

Corollary 4.6.8 There is a group $G$ of order $n^3$ given by $G= \{b^ic^ja^k \mid 0 ≤ i, j, k < n\}$, where $a$, $b$, and $c$ all have order $n$, and $b$ commutes with $c$, $a$ commutes with $c$, and $aba^{−1} = bc$. Thus, $(b^ic^ja^k) (b^{i'}c^{j'}a^{k'}) = b^{i+i'}c^{j+j'+ki'}a^{k+k'}$

The order of $(b^ic^ja^k)$ is n if n is odd and 2n if n is even.

Now suppose that $s: K \rightarrow H \rtimes_{\alpha} K$ is given by s(k)=(e,k) and $i : H \rightarrow H \rtimes_{\alpha} K$ is given by $i(h)=(h,e)$.

Proposition 4.6.9 Let α : K → Aut(H) be a homomorphism. Suppose given homomorphisms $f_1$ : H → G and $f_2$ : K → G for a group G. Then there is a homomorphism f : $H \rtimes_{\alpha}$ K → G with $f_1 = f$ ◦ i and $f_2$ = f ◦ s if and only if $f_2(k)f_1(h)(f_2(k))^{−1} = f_1(\alpha(k)(h))$ for all h ∈ H and k ∈ K.

Such an f, if it exists, is unique, and is given in the HK notation by $f(hk) = f_1(h)f_2(k)$.

a) In this problem we assume a familiarity with the group of invertible 3×3 matrices over the ring $Z_n$. Let

$$G = \left\{ \begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix} \mid \bar{x},\bar{y},\bar{z} \in Z_n \right\}$$

Show that $G$ forms a group under matrix multiplication. Show, using Proposition 4.6.9, that $G$ is isomorphic to the group constructed in Corollary 4.6.8.

I already showed that $G$ is a group under matrix multiplication, but I'm having trouble with the second part of this problem.

Let $G = \left\{ \begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix} \mid \bar{x}, \bar{y}, \bar{z} \in Z_n \right\}$

Let $G' = \{ b^ic^ja^k \mid 0 \leq i, j, k < n \}$ where a, b, c all have order n, and b commutes with c, a commutes with c, and $aba^{-1} = bc$. Thus,

$$(b^ic^ja^k)(b^{i'}c^{j'}a^{k'}) = b^{i+i'}c^{j + j' + ki'}a^{k+k'}.$$

I want to show that G is isomorphic to G' by using proposition 4.6.9

However, in order to use that proposition, I need to come up with an H and a K, right?

So...

Let $H = \left\{ \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} \right\}$

Let K=G

Obviously $G \cong H \rtimes_{\alpha} K$

Let $f_1 : H \rightarrow G’$ be the trivial homomorphism.

Let $f_2: K \rightarrow G'$ be given by $f_2 (\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}) = b^ic^ja^k$

Now we need to show that $f_2(\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix})f_1( \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix})f_2(\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{-1})$ = $f_1(\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{-1})$

But,

$$f_2(\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix})f_1( \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix})f_2(\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{-1}) = (b^ic^ja^k)(e)(b^ic^ja^k)^{-1} = e$$

and

$$f_1(\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{-1}) = e$$

So they must be equal.

So the homomorphism $f: H \rtimes_{\alpha} K \rightarrow G$ exists.

So now we just need to show that this homomorphism is an isomorphism.

I tried to show that by proving that $\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}$ has order n when n is odd and order 2n when n is even.

I first tried to show the case when n is odd:

Let n=2k+1 where $k \geq 0$

Base Case: n=0 is trivial.

We can also try n=3,

$( \begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix})^3 = \begin{bmatrix} 1 & 3(\bar{x}) & 3(\bar{y}) + 3(\bar{xy}) \\0 & 1 & 3(\bar{z}) \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}$

For the induction step, suppose for some 2w+1, an exponent of $\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}$ is 2w+1.

We need to show that, for 2(w+1)+1 = 2w+3, an exponent of $\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}$ must be 2w+3.

By induction, we know that

$$\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{2w+1} = \begin{bmatrix} 1 & (2w+1)\bar{x} & (2w+1)\bar{y}+(2w+1)\bar{xz} \\0 & 1 & (2w+1)\bar{z} \\0 & 0 & 1 \end{bmatrix}$$

However,

$$\begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{2w+3} = \begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^{2w+1} \begin{bmatrix} 1 & \bar{x} & \bar{y} \\0 & 1 & \bar{z} \\0 & 0 & 1 \end{bmatrix}^2 = \begin{bmatrix} 1 & (2w+1)\bar{x} & (2w+1)\bar{y}+(2w+1)\bar{xz} \\0 & 1 & (2w+1)\bar{z} \\0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2\bar{x} & 2(\bar{y})+\bar{xz} \\0 & 1 & 2\bar{z} \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2(\bar{x})+(2w+1)(\bar{x}) & 2(\bar{y})+\bar{xz}+(2w+1)(\bar{x})(2)(\bar{z})+ (2w+1)(\bar{xz})+(2w+1)\bar{y} \\0 & 1 & (2w+1)\bar{z}+2(\bar{z}) \\0 & 0 & 1 \end{bmatrix}$$

But that wasn’t the result we were supposed to get, right? Because $2(\bar{y})+\bar{xz}+(2w+1)(\bar{x})(2)(\bar{z})+ (2w+1)(\bar{xz})+(2w+1)\bar{y} = (2w+3)\bar{y} +(6w+4)\bar{xz}$ and 6w+4 is not a multiple of 2w+3.

So I was a bit confused…and I was wondering if anybody could help me with this.

$H,K$ and $\alpha$ need to be specified before we can talk about $H\rtimes_\alpha K$, so the bit between Corollary 4.6.8 and Proposition 4.6.9 is nonsensical. You seem to be equivocating between $G$ and $G'$ in the section where you're defining $f_1,f_2$, and I presume that you intended to have $K=G'$, yes? Also, your map $f_2$ doesn't seem to me to be clearly defined. What do $i,j,k$ have to do with $\overline x,\overline y,\overline z$? Then there's your meaningless sentence fragment "Furthermore, since the elements of $G$ and $G'$." Please try to address these issues, to aid understanding. – Cameron Buie May 19 '13 at 19:26
I expect they want you to take $H$ to be the subgroup with $x=0$ and $K$ to be the subgroup with $y=z=0$. But I would find this much easier to solve directly, without involving semidirect products. Just map $b^ic^ja^k$ to the matrix with $z=i$, $y=j$, $x=k$. – Derek Holt May 20 '13 at 7:46
@DerekHolt Thanks a lot. But the proposition tells us that there is a homomorphism $f:G \rightarrow G'$, right? So isn't that supposed to be an isomorphism? So why doesn't mine work? Did I do something worng while I was using the proposition? – user58289 May 20 '13 at 8:40