# Constructing Groups with Semi-direct product type question

1. Mar 8, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
e) If H ∼= Z3 × Z3 show that there are exactly 2 conjugacy classes of elements of order 2 in Aut(Z3 × Z3) = GL(2, Z3).

f) Choosing an element of each conjugacy class in e), construct two semidirect products of H and K. By counting orders of elements in each such group, show that the two groups you construct are not isomorphic.

2. Relevant equations

3. The attempt at a solution
So far I just have thoughts really...

As for finding the two conjugacy classes of GL(2,Z3) it seems that one of these conjugacy classes must be the identity matrix, because the identity matrix conjugated by anything will just return itself. So does one of the two conjugacy classes have the identity matrix and the other conjugacy class have everything else? Wouldn't these conjugacy classes have equal number of elements in each of them though? I am quite confused.

I am quite stuck. If anybody could shed some light on this topic for me that'd be awesome, I'll be editing this post as I think of new ideas.

2. Mar 8, 2016

### andrewkirk

That would not be the case because, if I am understanding the question correctly, it is asking about a partition of the set of elements of order two, and that set does not include the identity (and hence is a subset, but not a subgroup).

My interpretation is that the question is asking the following:

Given the definition $S\equiv\{a\in GL(2,\mathbb{Z}_3)\ |\ a^2=1\}$, prove that

$$\exists a,b\in S:\left[a\neq b\wedge \forall c\in S\exists g\in GL(2,\mathbb{Z}_3)\left( g^{-1}cg=a\vee g^{-1}cg=b\right)\right]$$

I suspect the answer may have something to do with the following.
$\mathbb{Z}_3$ has one automorphism of order two, which is the one that swaps 1 and 2. Hence I would guess that representative elements of the two desired conjugacy classes in Aut$(\mathbb{Z}_3\times\mathbb{Z}_3)$ would be
• the automorphism that swaps 1 and 2 in the first copy of $\mathbb{Z}_3$
• the automorphism that swaps 1 and 2 in the second copy of $\mathbb{Z}_3$

Last edited: Mar 8, 2016
3. Mar 9, 2016

### PsychonautQQ

Awesome, you're response was very insightful. This algebra stuff is tricky for me. Do you have any idea how once we've constructed the SDP using an element from each conjugacy class that we count the elements in the resulting group to show they are not isomorphic?

4. Mar 9, 2016

### andrewkirk

What is the semi-direct product that is to be constructed?
The question in the OP says it is from H and K, but it doesn't say what K is.

5. Mar 10, 2016

### PsychonautQQ

Oops. K is a sylow 2-subgroup of G. Also, H is a sylow 3-subgroup.

6. Mar 10, 2016

### PsychonautQQ

I'm sorry for my incompetance. G is a group of order 18 and K is a sylow 2-subgroup and H a sylow 3-subgroup (obviously).

I'm trying to construct a SDP noting that H is normal in G. In mapping K --> Aut(H) = Aut(Z3xZ3) there are two elements that divide |K|, the automorphism that permutes 1 and 2 in one of the Z3's and the autmorphism that does the same in the other Z3. We can now see that there are two conjugacy classes for elements of order 2 (because the elements of order 2 come in ordered pairs, and each one has one of the numbers in the pair the identity element).

Why does the question say to look at GL(2,Z3). Also, how can I learn enough about the SDP's formed by these morphisms to show that they don't create isomorphic groups?

7. Mar 10, 2016

### andrewkirk

We have |K|=2, because if K is a Sylow 2-subgroup of G and |G|=18=$2\cdot 3^2$ then every Sylow 2-subgroup of G has order 2. If that's correct then K has only one element $k$ aside from the identity $1_G$. Since we want each mapping $\phi_j:K\to Aut(H)$ to be a homomorphism, the order of $\phi(k)$ must be two, so we have two automorphisms to choose. Let $\phi_j$ map $k$ to the automorphism $\theta_j$ that swaps 1<-->2 in the $j$th coordinate of an element of $H$, for $j\in\{1,2\}$.

Then $P_j$, the $j$th semidirect product - ie the one based on $\phi_j$ - is the group made up of the cartesian product set $H\times K$, with multiplication defined by
$$(h_1,k_1)*(h_2,k_2)\equiv (h_1(\phi_j(k_1)(h_2)),k_1k_2)$$
for $j\in\{1,2\}$, and the identity element is $(1_G,1_G)$.

Our task is to prove that $P_1\not\cong P_2$.

Sounds tricky, as I would have expected them to be isomorphic. But if the question is correctly posed then, when we work it through, we'll find they are not.
Gotta dash now. I'll check back later.

Last edited: Mar 11, 2016
8. Mar 10, 2016

### PsychonautQQ

So we are creating the semi direct product by mapping K--->Aut(H). Since Aut(H) has two elements of order two, (one from each of the copies of Z3), there are two different morphisms we can use to build our SDP in H and K. How would the multiplication in our SDP's look diffferent depending on which of the two morphisms we use? Why would they be different at all? Also, why do you think my teacher said that Aut(H) = GL(n,Z3)? Do you think there is a way we can use that information possibly?

9. Mar 11, 2016

### andrewkirk

Yes I think that provides a useful way to approach the problem, because it provides a framework in which to find automorphisms. For instance the automorphism that swaps 1 and 2 in the first component has matrix
$$S_{1,2}\equiv\left(\begin{smallmatrix} 2&0\\ 0&1 \end{smallmatrix}\right)$$
and swapping in the second component has matrix
$$S^{1,2}\equiv\left(\begin{smallmatrix} 1&0\\ 0&2 \end{smallmatrix}\right)$$
Swapping in both is
$$S_{1,2}^{1,2}\equiv\left(\begin{smallmatrix} 2&0\\ 0&2 \end{smallmatrix}\right)$$

I'm now thinking my guess in post 2 about the representatives of the two conjugacy classes is wrong. $S_{1,2}$ and $S^{1,2}$ are too similar. I suspect there will be a matrix that can conjugate one into the other. What I wasn't taking into account is that, when we take the Cartesian Product of $\mathbb{Z}_3$ with itself, we introduce a new dimension for automorphisms, which is swaps between the two components. The automorphism of $H$ that swaps the two components has matrix
$$M\equiv\left(\begin{smallmatrix} 0&1\\ 1&0 \end{smallmatrix}\right)$$
My intuition now is that representatives of the two conjugacy classes would be $S^{1,2}$ and $M$, because they are very different from one another.

But first we need to prove that there are only two conjugacy classes. And before that I'd like to know whether there are any automorphisms of orders other than 2. We could look for an automorphism $T$ of order 3 by writing

$$T^3=\left(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\right)^3=\left(\begin{smallmatrix} 1&0\\ 0&1 \end{smallmatrix}\right)$$
and solving for $a,b,c,d$.

By the way, there may be theorems that can help with all this and provide speedy shortcuts to the solutions: theorems to do with Sylow subgroups, semi-direct products and automorphism groups. I haven't done any work in this area for quite a long time so I'm approaching it mostly from first principles.

EDIT: Hmm, with the benefit of a bit of quick R coding, here's a list of the matrices of four of the eight automorphisms of order three:

[,1] [,2]
[1,] 0 2
[2,] 1 2
[,1] [,2]
[1,] 1 1
[2,] 0 1
[,1] [,2]
[1,] 1 2
[2,] 0 1
[,1] [,2]
[1,] 2 2
[2,] 1 0

The other four are their transposes.

More later.

Last edited: Mar 11, 2016
10. Mar 12, 2016

### andrewkirk

A little programming can tell us that the order-two elements of $GL(2,\mathbb{Z}_3)$ are as per the following output
[1] "seeking automorphisms of order 2"
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 0 1
[2,] 1 0
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 0 2
[2,] 2 0
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 1 0
[2,] 0 2
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 1 1
[2,] 0 2
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 1 2
[2,] 0 2
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 1 0
[2,] 1 2
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 1 0
[2,] 2 2
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 2 0
[2,] 0 1
[1] "here is an automorphism with determinant 1"
[,1] [,2]
[1,] 2 0
[2,] 0 2
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 2 1
[2,] 0 1
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 2 2
[2,] 0 1
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 2 0
[2,] 1 1
[1] "here is an automorphism with determinant 2"
[,1] [,2]
[1,] 2 0
[2,] 2 1
[1] "number of automorphisms of this order found is 13"

Only one of them, the matrix $N\equiv \left(\begin{smallmatrix} 2&0\\ 0&2 \end{smallmatrix}\right)$, has determinant 1, and in $GL(2,\mathbb{Z}_3)$, conjugation cannot change the determinant. So that matrix must be in a conjugacy class of its own. That's the automorphism that swaps 2 with 1 in both the first and second coordinates of an element of $\mathbb{Z}_3\times\mathbb{Z}_3$ ($N$ is for negate, because 2=-1 in mod 3).

Hence, if there are only two conjugacy classes (result (e) - yet to be proved), then they are not symmetrical, as one will have twelve members and one will have only one. We can take as representative of the larger class the matrix $F\equiv \left(\begin{smallmatrix} 0&1\\ 1&0 \end{smallmatrix}\right)$, which swaps the first and second coordinates ($F$ is for flip).

So we can define our two SDPs respectively based on the homomorphisms that map the non-identity element $k$ of $K$ to $N$ and $F$ respectively.

Thus in SDP1 our multiplication rule will be
$$(h_1,k_1)*_1(h_2,k_2)\equiv (h_1N(h_2),k_1k_2)$$
if $k_1=k$ (ie is not the identity), and
$$(h_1,k_1)*_1(h_2,k_2)\equiv (h_1h_2,k_1k_2)$$
otherwise.

In SDP2 our multiplication rule will be
$$(h_1,k_1)*_2(h_2,k_2)\equiv (h_1F(h_2),k_1k_2)$$
if $k_1=k$, and the same as in SDP1 otherwise.

Write an element of either SDP as $((a,b),c)$ where $a,b\in\mathbb{Z}_3$ and $c\in K\equiv\{1,k\}$. Then our multiplication rules for the two SDPs, where the $K$ coordinate of the first operand is $k$ are

$((a,b),k)*_1((A,B),c)\equiv ((a,b)-(A,B),kc)$ for SDP1; and
$((a,b),k)*_1((A,B),c)\equiv ((a,b)+(B,A),kc)$ for SDP2.

Then investigate the order of elements of the form $((a,b),k)$. You will find that the number of those elements that have order two differs between the two SDPs. Putting that together with the observation that powers of elements of the form $((a,b),1_K)$ are the same in both SDPs (why?), this shows that the SDPs have different numbers of elements of order two, and hence cannot be isomorphic.

Last edited: Mar 12, 2016