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(a) What are the finite abelian groups of order 100 up to isomorphism?

(b) Say $G$ is a finite abelian group of order 100 which contains an element of order 20 and no element with larger order. Then G will be isomorphic to exactly one group from your list in (a): which one, and why?

Can someone tell me how to analyze this question? As for question (a), I know that finite abelian group could be isomorphic to $Z_{a_1}\times Z_{a_2}\times...\times Z_{a_n}$, but have no idea what's the relation between $a_1,a_2,...,a_n$ and 100. Can someone tell me how to analyze this question?

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  • $\begingroup$ Hint: what's the order of the group $Z_2 \times Z_5$? That order depends on $2$ and $5$ in a simple way. $\endgroup$ – Hew Wolff May 7 '15 at 2:02
  • $\begingroup$ HINT: Lagrange's Theorem states that the order (that is to say number of elements) of a subgroup $H$ of a group $G$ divides the order of $G$. i.e. If $H \leq G \Rightarrow \frac{|G|}{|H|} \in \mathbb{N}$ $\endgroup$ – Krishan Bhalla May 7 '15 at 2:03
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We just have to use the fundamental theorem for finite abelian groups.

Note $100=2^2\cdot 5^2$

You just have to choose a factorization of $2^2$ and a factorization of $5^2$.

The only two factorizations for $2^2$ are $2\cdot 2$ and $4$

the only two factorizations for $5^2$ are $5\cdot 5 $ and $25$.

So there are $4$ combinations, these give us all the groups:

$\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_5\times \mathbb Z_5$

$\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_{25}$

$\mathbb Z_4\times \mathbb Z_{5}\times \mathbb Z_5$

$\mathbb Z_4\times \mathbb Z_{25} \cong \mathbb Z_{100}$


The only of these that has maximum order $20$ is the third element in the list.

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  • $\begingroup$ a question:does every finite abelian's order equal to a1*a2*.......*an,if the finite abelian group is isomorphic to $Z_{a1}\times Z{a2}...\times Z_{an}$? $\endgroup$ – python3 May 7 '15 at 2:11
  • $\begingroup$ Yes, that is what the fundamental finite abelian group theorem says, you just have to figure out you can factor every prime power factor, in this case the prime power factors are $2^2$ and $5^2$, so there are not many ways to do so. $\endgroup$ – Jorge Fernández Hidalgo May 7 '15 at 2:12
  • $\begingroup$ Oh, thanks,I got it $\endgroup$ – python3 May 7 '15 at 2:13

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