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Homomorphisms with unknown groups 
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#1
Feb1913, 01:23 PM

PF Gold
P: 2,281

1. The problem statement, all variables and given/known data
1)Let p,q be primes. Show that the only group homomorphism $$\phi: C_p \mapsto C_q$$ is the trivial one (i.e ## \phi (g) = e = e_H\,\forall\,g##) 2)Consider the function $$det: GL(n,k) \mapsto k^*.$$ Show that it is a group homomorphism and identify the kernel and image. 3. The attempt at a solution I cannot make much progress with either of these questions essentially because I do not know what the groups are. I don't know what ##C_p## or ##C_q## are. Any ideas? For the second one, I can at least make a start: Take A,B in GL(n,k). GL(n,k) is a group under multiplication of matrices, so det(g1 * g2) = det(AB) = (detA)(detB) from linear algebra = det(g1)* det(g2) since det(gi) ##\neq## 0 for all gi. Again, what is ##k^*##? From the homomorphism I just showed, it is probably a group under multiplication of numbers and GL(n,k) = set of invertible nxn matrices with real number entries, but I want to make sure. So, Ker (det) = {##g \in GL(n,k) \mid det(g) = e_H = 1##} = SL(n,k) Im (det) = {##h \in k^* \mid \exists g \in GL(n,k) , det(g) = h##} = ##k^* ##\## {0}##. I don't know what ##k^*## is so whether I need to exclude 0 or not, I don't know. Many thanks. 


#2
Feb1913, 01:50 PM

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PF Gold
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#3
Feb1913, 01:53 PM

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Hmm for the first one, I THINK they might be modulo groups. Contradiction is a good approach to it. That is, assume that phi is not the only group homomorphism and then proceed to show that phi is indeed the only one.



#4
Feb1913, 01:57 PM

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PF Gold
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Homomorphisms with unknown groups
Your kernel looks fine, and so does your image. (Note that ##k^* \setminus \{0\} = k^*##.) You may want to give an specific example to show that for any ##h \in k^*##, you can find a matrix ##A## with ##det(A) = h##. 


#5
Feb1913, 02:21 PM

PF Gold
P: 2,281

So assuming they are cyclic groups,I have ##Ker \phi = \left\{g^{nq}\right\}, n \in \mathbb{N}, q## the order of ##C_q## and ## Im \phi = \left\{e_H\right\}## because the element in Ker phi will give the identity. I don't know if I can say for sure if there exists more elements to Im phi. Does this help me?



#6
Feb1913, 02:31 PM

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PF Gold
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Instead of trying to construct the kernel explicitly, consider what its order can be. (Use Lagrange's theorem.) Do the same thing for the image. What are the possibilities? 


#7
Feb1913, 02:42 PM

PF Gold
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#8
Feb1913, 02:46 PM

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If ##ker \phi = p##, then what is ##ker \phi##, and what is ##\phi##? 


#9
Feb1913, 02:50 PM

PF Gold
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#10
Feb1913, 02:55 PM

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So now all we have to do is rule out the other possibility, that ##ker \phi = 1##. If this were true, what does it imply about the map ##\phi##? Injective, surjective, something like that? 


#11
Feb1913, 03:06 PM

PF Gold
P: 2,281

But if we have injectivity, then order of g in G = order of ##\phi(g)## in H. By assumption, p and q were different and so this is a contradiction. Hence, Ker \phi = 1 is ruled out. EDIT: Question should say 'p,q are different primes' 


#12
Feb1913, 03:32 PM

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#13
Feb1913, 03:37 PM

Mentor
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It would be an interesting exercise to see how many homomorphisms [itex]\varphi:C_p\rightarrow C_p[/itex] there are.
Since they are all bijections except for the trivial map, we can look at the set of all bijective homomorphisms. This forms a group. It would be a good exercise to find out which group it is (answer: it will be the cyclic group on p1 elements). 


#14
Feb1913, 04:01 PM

PF Gold
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#15
Feb1913, 04:32 PM

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If we wanted to prove it directly, we could suppose that ##\phi## is a surjective homomorphism. Then ##im(\phi) = q##. But by the first isomorphism theorem, we have ##C_p / ker(\phi) \cong im(\phi)##, so ##im(\phi)## must divide ##p##. This is a contradiction. 


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