# groups homomorphisms

• Feb 2nd 2008, 01:50 PM
joanne_q
groups homomorphisms
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.
• Feb 2nd 2008, 03:52 PM
ThePerfectHacker
Quote:

Originally Posted by joanne_q
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 $\varphi$

The second part of the question is asking to find the inverse for $(k,q)$. It is an element $(x,y)$ such that $(k,q)(x,y) = 1 = (x,y)(k,q)$. Thus, $(k \phi (q) x,qy) = (1,1)$. Which means $qy = 1 \implies q = y^{-1}$. And $k\phi (q) x = 1 \implies x =k^{-1} \phi^{-1} (q)$.
• Feb 2nd 2008, 06:47 PM
joanne_q
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 $k\phi (q_1) x = 1 \implies x =k^{-1} \phi^{-1} (q_1)$
• Feb 3rd 2008, 02:34 PM
joanne_q