Distinct composition series of $Z_{30}$

Question

Given two composition series for a group G: $$\{e \} \triangleleft G_1 \triangleleft ... \triangleleft G_k \triangleleft G $$ $$\{e \} \triangleleft H_1 \triangleleft ... \triangleleft H_k \triangleleft H $$

If we say they are distinct when $G_i \ncong H_i$ for at least one i. How many distinct composition series does $Z_{30}$ have?

I am a bit confused with this question I initially thought it was linked to the Jordan-Holder theorem and so there would only be 1, but then I realized that this only tells us that series only have the same composition factors up to isomorphism.

I just wanted to check that I am right and this doesn't hold for individual $G_i \cong H_i$ too?

If this is true does anyone have any hints for an easy way to work this out or is it just a case of trying to find all the normal subgroups of $Z_{30}$, and would Sylow theorems possibly help to find these?

Answer 1

You should be able to find all the normal subgroups of $\Bbb Z_{30}$ without Sylow. The group is abelian, and therefore all subgroups are normal. The group is cyclic, so all subgroups are cyclic.

As for how to make different composition series, consider the quotients $G_{i+1}/G_i$. In your case, the three groups that will appear as those quotients will necessarily be $\Bbb Z_2, \Bbb Z_3$ and $\Bbb Z_5$. But you can make them appear in any order.

Answer 2

You already have a good answer. It might also be useful to know (you can probably do this as an easy exercise now that you understand that order 30 case) that a finite abelian group has only one composition series if and only if it has order the power of a prime.

Answer 3

$Z_{30}={0,1,2,...29}$ (addition $\bmod 30$ here):

$\langle2\rangle= {0,2,4,...28}$
$\langle3\rangle= {0,3,6,...,27}$ and
$\langle5\rangle$, $\langle6\rangle$, $\langle10\rangle$, $\langle15\rangle$ are subgroups of $Z_{30}$ other then ${0}$ and $Z_{30}$.

Therefore, composition series are
${0}⊂\langle15\rangle⊂\langle5\rangle⊂G$
${0}⊂\langle15\rangle⊂\langle3\rangle⊂G$
${0}⊂\langle10\rangle⊂\langle5\rangle⊂G$
${0}⊂\langle10\rangle⊂\langle2\rangle⊂G$
${0}⊂\langle6\rangle⊂\langle3\rangle⊂G$
${0}⊂\langle6\rangle⊂\langle2\rangle⊂G$

for the first series is $\langle5\rangle/\langle15\rangle$ and $|\langle5\rangle/\langle15\rangle|= |\langle5\rangle|/|\langle15\rangle|=6/2=3$ which is prime no.

therefore $\langle5\rangle/\langle15\rangle$ is simple similarly, we can show that all factor group $G_i/G_{i+1}$ are simple.