Show Quotient Groups are isomorphic

[tehme2]

Homework Statement

Show that Z18/M isomorphic to Z6 where m is the cyclic subgroup <6>
operation is addition

The Attempt at a Solution

M = <6> , so M = {6, 12, 0}

I figured I could show that Z18/M has 6 distinct right cosets if I wanted to do M + 0 = {6, 12, 0} M + 1 = {7, 13, 1} M + 2 = {8, 14, 2} etc and after M + 5 , they would start to repeat

After that I do not know how you show isomorphism. I know it has to satisfy three conditions, injective, surjective, and homomorphism, but I do not know what the function is even, so I'm assuming it is f(a) = M + a for a an element of Z18

so injective : if f(a) = f(b) so ( M +a = M + b ), then a = b.
M + a = {6 +a, 12 + a, a} and M + b = {6 + b, 12 + b, b}
then 6 + a = 6 + b
12 + a = 12 + b
a = b , so I guess I need to show that each of those three equations show a = b, and I will use cancellation of 6, and 12 and we get a = b, so f is injective (but I don't understand how it is injective

surjective: let a be arbitrary element of Z18, then f(a) = M + a = {6 + a, 13 + a, a} so a would be in Z18/M , so it is surjective

homomorphism: f(a + b) = f(a) + f(b) , so
M + ( a + b) = M + a + M + b = f(a) + f(b) this result came from definition of a quotient group from my book which said that G/N (N a normal subgroup) is a group under the operation defined by(Na)(Nc) = N(ac)


Now my book gave a hint saying show that 1 +M generates all of Z18/M, and so Z18/M is cyclic and then show that the order of (1 + M) is 6, however, I do not quite understand how you would show that a set has order since 1 + M = {7, 13, 1} right?
much less showing that 1 + M generates all of Z18/M. thanks

 

[gopher_p]

There are three groups that you're looking at here; $Z_{18}$, $Z_6$, and the quotient group $Z_{18}/M$.

Which of these two groups are you meant to show are isomorphic? In your work, what is the "domain" of the isomorphism that you're trying to construct?

If I were you, I would focus on understanding the quotient group a little better before you try to solve the problem. What are the elements of the quotient group? Does an element of the quotient group have multiple names? Can you make a Cayley table for this group? Can you find a homomorphism from $Z_{18}$ onto $Z_{18}/M$? what is the kernel of this homomorphism?

 

[tehme2]

I'm meant to show the quotient group Z18/M is isomorphic to Z6.

The elements of Z18/M are
M + 0 = {0, 6, 12}
M + 1 = {1, 7, 13}
M + 2 = {2, 8, 14}
M + 3 = {3, 9, 14}
M + 4 = {4, 10, 15}
M + 5 = {5, 11, 16}

so, I need a function f that takes z18/m to Z6 , so the domain would be the set of Z18/M which are the elements above.

The elements of the quotient group can only have one name, since the quotient group is the set of distinct right cosets.

Yes a Cayley table shows that M + 0 is 0
M + 1 is 1
M + 2 is 2
M + 3 is 3
M + 4 is 4
M + 5 is 5
where 0,1, 2, 3, 5, would be the elements from Z6
I've attached the Cayley table I made.

I'm sure we could find a homomorphism from Z18 onto Z18/M
f : Z18 → Z18/M where f(a) = M + a
then f(a+b) = M + (a+b) = (M +a) + (M + b) = f(a) + f(b)
the kernel would be the set of elements in Z18 that get mapped to the identity element of Z18/M which is the coset M
Then the kernel is {0, 6, 12}


so I suppose since we need to show Z18/M is isomorphic to Z6 we could use the Cayley tables since it would seem each element of Z18/M corresponds to one element from Z6 and we basically just relabel the table of Z18/M with the elements they correspond to from Z6 and we have the same tables.

 

[gopher_p]

Ok, good. It looks like you've found the correct mapping $\phi:\mathbb{Z}_{18}/M\rightarrow \mathbb{Z}_6$. It's bijective (which I think we can agree is obvious). It's also a homomorphism, which may not be as obvious. Can you show that $\phi$ is a homomorphism?

While I think you've accomplished the goal of the problem, it may not be exactly what the problem was supposed to help you understand. If I had to guess, you've recently covered the First Isomorphism Theorem? See if you can find a surjective homomorphism $\Phi:\mathbb{Z}_{18}\rightarrow \mathbb{Z}_6$ with $Ker(\Phi)=M$. Try to see how this would give you an alternate way to solve the problem. The various Isomorphism Theroems are a big deal, and it's helpful to see how they work in these "easy" cases so that we believe that they work in the more difficult cases.

 

[jgens]

If you have proven that $Z_6$ is the unique abelian group of order 6, then you can also just note that $Z_{18}/\langle 6 \rangle$ is an abelian group of order 6. This is another useful technique in finite group theory.