# Show Quotient Groups are isomorphic

## Homework Statement

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

## 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

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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?

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.

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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
Gold Member
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.