# Are these semidirect products of groups isomorphic?

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1. May 7, 2016

### pondzo

1. The problem statement, all variables and given/known data

Write $C_3\langle x|x^3=1\rangle$ and $C_2=\langle y|y^2=1\rangle$
Let $h_1,h_2:C_2\rightarrow \text{ Aut}(C_3\times C_3)$ be the following homomorphisms:
$$h_1(y)(x^a,x^b)=(x^{-a},x^{-b})~;~~~~~~h_2(y)(x^a,x^b)=(x^b,x^a)$$
Put $G(1)=(C_3\times C_3)\rtimes_{h_1}C_2 \text{ and } G(2)=(C_3\times C_3)\rtimes_{h_2}C_2$

Is $G(1)\cong G(2)$? If so write down an explicit isomorphism. If not, explain why not.

2. Relevant equations

I think the group operation for this semidirect products is:
$$\cdot ~:~(C_3\times C_3)\rtimes_{h_i}C_2~~~~~~~i=\{1,2\}$$ $$\text{ Where } ((x_1,x_2),y)\cdot ((x_1',x_2'),y')=((x_1,x_2)h(y)(x_1',x_2'),yy')$$ $$\text{ for } ((x_1,x_2),y),((x_1',x_2'),y')\in (C_3\times C_3)\rtimes_{h_i}C_2$$

3. The attempt at a solution

First of all, i'm not too sure if I have the elements in each group in the correct form or if I defined the group operation correctly. Assuming I have done these two things correctly, I will continue with my attempt of a solution.

Write $C_3\times C_3=\langle x_1,x_2|x_1^3=x_2^3=1, x_2x_1=x_1x_2\rangle$.
Let $Y=((1,1),y)$ and $X=((x_1,x_2),1)$
I will now do the crucial calculation, $YX$, in both G(1) and G(2):

Crucial calculation in $G(1)$:
$YX=((1,1),y)\cdot((x_1,x_2),1)=((1,1)h(y)(x_1,x_2),y1)=(((x_1)^{-1},(x_2)^{-1}),y)=(((x_1)^2,(x_2)^2),y)=X^2Y$
Thus $G(1)=\langle X,Y|X^3=Y^2=1,YX=X^2Y\rangle\cong D_6$ (Binary dihedral group of order 6)

Crucial calculation in $G(2)$:
$YX=((1,1),y)\cdot((x_1,x_2),1)=((1,1)h(y)(x_1,x_2),y1)=((x_1,x_2),y)=XY$
Thus $G(2)=\langle X,Y|X^3=Y^2=1,YX=XY\rangle \cong C_3\times C_2$

So $G(1)\ncong G(2)$.

I am not sure If my final conclusions about G(1) and G(2) are correct. I.e. i'm not sure if I can write $G(1)=\langle X,Y|X^3=Y^2=1,YX=X^2Y\rangle$ and $G(2)=\langle X,Y|X^3=Y^2=1,YX=XY\rangle$

2. May 7, 2016

### Staff: Mentor

I've checked everything and it is correct. G(1) and G(2) are correctly represented and identified.
(At least I've found nothing wrong and I even checked whether the normal subgroup is on the correct side of $\rtimes$.)

3. May 7, 2016

### micromass

Staff Emeritus
Aren't $G(1)$ and $G(2)$ supposed to have $18$ elements? Why do you suddenly conclude they have $6$ elements?

4. May 7, 2016

### pondzo

Upon further consideration I agree with you. I think my conclusions had something to do with how I represented the generators of the $(C_3\times C_3)$ part of the semidirect product. Would it be correct to denote $X_1=((x,1),1)$ and $X_2=((1,x),1)$ and then do the crucial calculations $YX_1$ and $YX_2$ in both G(1) and G(2)?

So, the crucial calculations in G(1) would be:
$YX_1=((1,1),y)\cdot((x,1),1)=((1,1)h(y)(x,1),y1)=((x^{-1},1),y)=((x^2,1),y)=X_1^2Y$
$YX_2=((1,1),y)\cdot((1,x),1)=((1,1)h(y)(1,x),y1)=((1,x^{-1}),y)=((1,x^2),y)=X_2^2Y$
Thus $G(1)\cong \langle X_1,X_2,Y|X_1^3=X_2^3=Y^2=1,YX_1=X_1^2Y,YX_2=X_2^2Y,X_2X_1=X_1X_2 \rangle\cong ??$ (Is there a name for this?)

And the crucial calculations in G(2) would be:
$YX_1=((1,1),y)\cdot((x,1),1)=((1,1)h(y)(x,1),y1)=((x,1),y)=X_1Y$
$YX_2=((1,1),y)\cdot((1,x),1)=((1,1)h(y)(1,x),y1)=((1,x),y)=X_2Y$
Thus $G(1)\cong \langle X_1,X_2,Y|X_1^3=X_2^3=Y^2=1,YX_1=X_1Y,YX_2=X_2Y,X_2X_1=X_1X_2 \rangle\cong C_3\times C_3 \times C_2$

And so $G(1)\ncong G(2)$ since one of these groups is abelian and the other isn't.

5. May 7, 2016

### Staff: Mentor

No. It's simply the direct product of the symmetric group $S_3 ≅ D_3$ with $C_3$. (You denoted $D_3$ as $D_6$.)

I would have concluded with the normal subgroups $<YX>$ you defined in the first place.
If $G(1)$ and $G(2)$ were isomorphic, you could consider the quotient groups and the fact there is only one group of order three.

Last edited: May 7, 2016