Are these semidirect products of groups isomorphic?

[pondzo]

Homework Statement

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.

Homework Equations

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

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##

 

[fresh_42]

Homework Statement

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.

Homework Equations

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

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##

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##.)

 

[micromass]

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

 

[pondzo]

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

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.

 

[fresh_42]

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?)

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##.)

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.

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.