# Group Homomorphisms: Verifying Group Property & Finding Inverse

• mathusers
In summary: In other words, for any two automorphisms \psi, \phi\in Aut(K), \psi\circ\phi\in Aut(K) also. so that is why \varphi(q_1q_2) = \varphi(q_1)\circ\varphi(q_2).
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.

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?

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'.

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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 $k\phi (q_1) x = 1 \implies x =k^{-1} \phi^{-1} (q_1)$

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

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

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??

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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.

quasar987 said:
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?

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.

## 1. What is a group homomorphism?

A group homomorphism is a function that preserves the algebraic structure of a group. In other words, it maps elements of one group to elements of another group in a way that respects the group operation.

## 2. How do you verify if a function is a group homomorphism?

To verify if a function is a group homomorphism, you need to check if it preserves the group operation. This means that for any two elements a and b in the group, the function must satisfy f(a * b) = f(a) * f(b). If this property holds, then the function is a group homomorphism.

## 3. What is the inverse of a group homomorphism?

The inverse of a group homomorphism is a function that "undoes" the original function, or maps elements back to their original group. This can be found by first finding the inverse function of the original function, and then verifying that it is also a group homomorphism.

## 4. How do you use group homomorphisms to solve problems?

Group homomorphisms can be used to simplify calculations and proofs in abstract algebra. For example, if we know that two groups are isomorphic, meaning there exists a group homomorphism between them, we can use this to show that they share many of the same properties and structures.

## 5. Can a group homomorphism have a non-trivial kernel?

Yes, a group homomorphism can have a non-trivial kernel. The kernel of a group homomorphism is the set of elements in the domain that are mapped to the identity element in the codomain. If the homomorphism is not injective, meaning different elements in the domain are mapped to the same element in the codomain, then the kernel will be non-trivial.

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