# Group homomorphisms

1. Feb 2, 2008

### mathusers

hi a little help would be kindly appreciated here guys.

any suggestions on how to go about doing these?

INFORMATION
-----------------------

if K,Q are groups $\varphi : Q \rightarrow Aut(K)$ is a homomorphism the semi direct product $K \rtimes_{\varphi} Q$ is defined as follows.

(i) as a set $K \rtimes_{\varphi} Q = K \times Q$
(ii) the group operation * is $(k_1,q_1)*(k_2,q_2) = (k_1 \varphi(q_1)(k_2),q1q2)$

THE QUESTION
-----------------------

Verify formally that $K \rtimes_{\varphi} Q = (K \times Q, *, (1,1)$ is a group and find a formula for $(k,q)^{-1}$ in terms of $k^{-1},q^{-1}$ and $\varphi$

-----> to show that it is a group, i know i have to show that the 4 conditions for being a group (e.g. associativity, closure, existance of identity element, existance of inverse) have to be satisfied. but not really too sure how to show it.. and im completely baffled for the 2nd part of the question.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 2, 2008

### Mystic998

I'm not sure what the (1, 1) at the end of the triple (quadruple?) is supposed to indicate, but the verification that the semidirect product actually forms a group is pretty simple. I'll do closure, and maybe you'll get the idea. As a note, I'm going to denote $\varphi(q_{1})$ by $\phi_{q_{1}}$ because that makes it a lot more clear that the image is actually an automorphism of K.

$(k_{1}, q_{1})\ast(k_{2}, q_{2}) = (k_{1}\phi_{q_{1}}(k_{2}), q_{1}q_{2})$ by definition. Since $\phi_{q_{1}}$ is an automorphism of K, $\phi_{q_{1}}(k_{2})=k_{3}$ for some $k_{3} \in K$. Since K is a group, $k_{1}k_{3} \in K$. Similarly, it's clear that $q_{1}q_{2} \in Q$. Hence $(k_{1}\phi_{q_{1}}(k_{2}), q_{1}q_{2}) \in K \times Q$, and closure holds.

Does that help?

3. Feb 2, 2008

### quasar987

The (1,1) at the end of the quadruple is there to emphasize the fact that the identity in the semi direct product is the couple (1_K,1_Q).

The most tedious part of checking the semi direct product is a group is...associativity! But it certainly is not difficult.

For the second part of the question, you are asked to find an expression for the inverse of a general element (k,q) in terms of the inverses of k and q and the function phi.

I'll do the easy half of the question for you. We want $(k',q')=(k,q)^{-1}$ such that $$(1,1)=(k,q)(k',q')=(\mbox{complicated expression},qq')$$

This implies that q' must be q^{-1}. Now all you gotta do is solve "complicated expression = 1" for k'.

Last edited: Feb 2, 2008
4. Feb 2, 2008

### mathusers

thnx a lot for the help..
here is my working out so far: please verify and suggest any corrections for wrong the parts.

Checking of the 4 conditions for a group
--------------------------------------

Closure:
$(k_1,q_1) * (k_2,q_2) = (k_1\varphi(q_1)(k_2), q_1q_2)$.
Since $\varphi(q_1)$ is an automorphism of K, $\varphi(q_1)(k_2) = k_3$ for some $k \epsilon K$. Since K is a group, $k_1,k_3 \epsilon K$. Similarly, $q_1,q_2 \epsilon Q$. So $(k_1\varphi(q_1)(k_2), q_1q_2) \epsilon K \times Q$ and closure holds.

Existance of identity element:
$(k_1,q_1) * (1,1) = (k_1\varphi(q_1)(1), q_1(1)) = (k_1,q_1)$ and
$(1,1) * (k_1,q_1) = (1\varphi(1)(k_1), 1(q_1)) = (k_1,q_1)$.
So $(k_1,q_1) * (1,1) = (1,1) * (k_1,q_1) = (k_1,q_1)$.
So the identity element exists.

Existance of inverse element:
The inverse is an element $(x,y)$ such that $(k_1,q_1)(x,y) = (k_1\varphi(q_1)x, q_1y) = (1,1)$ and
$(x,y)(k_1,q_1) = (x\varphi(y)k_1, yq_1) = (1,1)$.
so $(k_1,q_1)(x,y) = (1,1) = (x,y)(k_1,q_1)$

So the inverse element exists.

Associativity:
$[(k_1,q_1)*(k_2,q_2)] * (k_3,q_3) = [(k_1\varphi(q_1)(k_2), q_1q_2)] * (k_3,q_3) = (k_1\varphi(q_1)(k_2)\varphi(q_1q_2)(k_3), q_1q_2q_3)$

$(k_1,q_1) * [(k_2,q_2)*(k_3,q_3)] = (k_1,q1) * [(k_2\varphi(q_2)(k_3),q_2q_3)] = (k_1\varphi(q_1)k_2\varphi(q_2)(k_3), q_1q_2q_3)$

the 2 answers dont seem to match up here. please verify this.

2ND PART OF QUESTION

Finding the inverse for $(k_1,q_1)$.
The inverse is an element $(x,y)$ such that $(k_1,q_1)(x,y) = (1,1) = (x,y)(k_1,q_1)$. So $(k_1 \phi (q_1) x,q_1y) = (1,1)$. This means $q_1y = 1 \implies q_1 = y^{-1}$. And $k\phi (q_1) x = 1 \implies x =k^{-1} \phi^{-1} (q_1)$

Last edited: Feb 3, 2008
5. Feb 3, 2008

6. Feb 3, 2008

### quasar987

Concerning associativity, you still need to show using the properties of homomorphisms that

$$k_1\varphi(q_1)(k_2)\varphi(q_1q_2)(k_3) = k_1\varphi(q_1)(k_2\varphi(q_2)(k_3))$$

7. Feb 3, 2008

### mathusers

concerning associativity, thats exactly where im stuck at, where would i go from that line..

any hints?

i understand that $\varphi$ is a holomorphism but i dont know what $\varphi(q_2)$ or $\varphi(q_1q_2)$ represent or how i would interpret them??

Last edited: Feb 3, 2008
8. Feb 3, 2008

### quasar987

phi is itself a homomorphism that send element of Q to automorphisms of K!

so $\varphi$ is an homomorphism, and for any q in Q, $\varphi(q)$ is again a homomorphism.

9. Feb 3, 2008

### mathusers

thnx i understand that but what i mean is what values do $\varphi(q)$ and $\varphi(qq_2)$ take? visibly they must be the same since the 2 overall equations have to be the same for associativity to hold?

10. Feb 3, 2008

### quasar987

The point is that phi and phi(q) are both homomorphisms is that the two expressions can be simplified. Let me do the first one to demonstrate the idea...

$$k_1\varphi(q_1)(k_2)\varphi(q_1q_2)(k_3)=k_1\varphi(q_1)(k_2)[\varphi(q_1)\circ \varphi(q_2)](k3)=k_1\varphi(q_1)(k_2)\varphi(q_1)(\varphi(q_2)(k3))$$

notice the "$\circ$" in the second step? this is because the group operation in Aut(K) is the composition of function.