Group Homomorphisms: Verifying Group Property & Finding Inverse

[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, existence of identity element, existence of inverse) have to be satisfied. but not really too sure how to show it.. and I am completely baffled for the 2nd part of the question.
please help out :) thnx

 

[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?

 

[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 got to do is solve "complicated expression = 1" for k'.

 

[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 don't 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 $k\phi (q_1) x = 1 \implies x =k^{-1} \phi^{-1} (q_1)$

 

[mathusers]

any corrections please?

 

[quasar987]

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

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

 

[mathusers]

concerning associativity, that's exactly where I am stuck at, where would i go from that line..

any hints?

i understand that $\varphi$ is a holomorphism but i don't know what $\varphi(q_2)$ or $\varphi(q_1q_2)$ represent or how i would interpret them??

 

[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.

 

[mathusers]

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.

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?

 

[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.