Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the semidirect products of $C_2 \times C_2$ by $C_3$, that is: $(C_2 \times C_2) \rtimes C_3$

My approach: I let $C_3:=\langle y\rangle$ and let $\phi : C_3 \to \mathrm{Aut}(C_2 \times C_2)$ be a homomorphism, observe that $\mathrm{Aut}(C_2 \times C_2) \cong \mathrm{GL}(2,2)$ and $\phi(y)$ has order 1 or 3.

Case 1: If $\phi(y)$ has order 1, then $\phi$ is the trivial homomorphism thus $(C_2 \times C_2) \rtimes C_3 \cong C_2 \times C_2 \times C_3 $

Case2: If $\phi(y)$ has order 3, then $\phi(y)= \begin{pmatrix}0 & 1\\ 1 & 1\end{pmatrix} $ which is the only matrix of order 3 inside $\mathrm{GL}(2,2)$, in the next step one needs to find the isomorphism type of $(C_2 \times C_2) \rtimes C_3$ relised by $\phi$ in this case, this is where my problem is: I know its presentation $\langle a,b,y\rangle$ contains $ \{ a^2=1,\ b^2=1,\ [a,b]=1,\ y^3=1 \} $ as a part of the relations, but I do not know how to get all the relations, and also how to get its isomorphism type.

share|improve this question
1  
I guess the cheap way to show that this is $A_4$ is to note that it's the only non-abelian group of order $12$ with a unique subgroup of order $4$. –  Dylan Moreland May 31 '12 at 4:15
1  
Note that you used different notation for cyclic groups in the title and in the body. Moreover, in the title you had $Z_2*Z_2$; that's the free product, not the direct product. –  Arturo Magidin May 31 '12 at 4:16
1  
Isn't the idea that conjugation by $x$ gives the automorphism $\phi(y)$? Also, instead of matrices you can think of $\phi(y)$ as simply permuting the three non-trivial elements $a,b,ab$ of $C_2\times C_2$ in a 3-cycle. So $xax^{-1}=b$, $xbx^{-1}=ab$, $\ldots$ Some redundancy may already be here, –  Jyrki Lahtonen May 31 '12 at 4:18
    
Also, if there's one matrix of order $3$ inside $GL_2(\mathbf Z/2\mathbf Z)$ then there are two. [Look at the inverse of the one you have.] This doesn't give you a new semidirect product, though. –  Dylan Moreland May 31 '12 at 4:19
1  
Your presentation needs to say how $x$ interacts with $a$ and $b$; otherwise, what you have is $(C_2\times C_2)*C_3$, the free product of $C_2\times C_2$ with $C_3$. The missing relations are $a^x = b$ and $b^x = ab$. –  Arturo Magidin May 31 '12 at 4:21
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.