# An abelian group of order 100

The first part of the problem asks you to prove that an abelian group $G$ with order $100$ must contain an element of order $10$. For this part, I use Sylow theorem to list possiblities for $H$ and $K$ where $|H|$=$2^{2}$ and $|K|$=$5^{2}$. $K$ must be normal since there is not subgroup of order $25$ while $H$ might not be normal since $1$$\equiv$$25$ mod $2$. Also I proved that H$$K=G and they have only the identity element in common. Then, if H and K are normal. G must be isomorphic to one of the following groups: Z_{4} \times Z_{25} Z_{2} \times Z_{2} \times Z_{25} Z_{4} \times Z_{5} \times Z_{5} Z_{2} \times Z_{2} \times Z_{5} \times Z_{5} From above, it is easily to pick up element of order 10 for each. But my confusion is that since G is an abelian group, how can I use the theorem that any finite abelian group is isomorphic to a direct product of cyclic groups? Also, since the second part asks you if no element of G has order greater than 10, what are its torsion coefficients? I think my way of listing the possibilities are too complicated. Are there any more explicit ways to solve the problem? Thanks a lot. - Is your overarching assumption that G is abelian? Or do you need to prove G is abelian after the first part of the problem? – Arturo Magidin May 10 '12 at 18:50 @ArturoMagidin Sorry for the confusion, it is already given that G is an abelian. That's why I am thinking of using the finite abelian group theorem but find the sylow theorem more easily to handle. – Simonaster May 10 '12 at 18:55 Gioven your group is commutative, can you find, by any means, an element of order 2, and an element of order 5? What are the possible orders of elements greater than 10 in an abelian group of order 100? Which orders must you have? How do you exclude these possibilities? – Mark Bennet May 10 '12 at 19:03 @Simonaster: I don't think you should ever need Sylow's theorems when working with finite abelian groups. Note also that every subgroup of an abelian group is normal. – Tara B May 10 '12 at 19:07 You don't need the structure theorem of finitely generated abelian groups to solve this problem. It suffices to have Cauchy and Lagrange in hand. – Arturo Magidin May 10 '12 at 19:20 ## 1 Answer If you are assuming that G is an abelian group of order 100, then you don't need Sylow's Theorems, you just need Cauchy's Theorem: since 2 and 5 divide |G|, G has an element a of order 2, and an element b of order 5. Then \langle a\rangle\cap\langle b\rangle = \{0\} (writing the groups additively), so a+b has order 10, as is easily verified:$$\begin{align*} k(a+b) = 0 &\iff ka+kb = 0\\ &\iff ka=-kb\\ &\iff ka,kb\in\{0\}\\ &\iff 2|k\text{ and }5|k\\ &\iff 10|k. \end{align*}$The second part asks you to whether there are no elements of order greater than 100 in an abelian group of order 100? The cyclic group of order 100 shows that this need not be the case. In fact, the only abelian group of order$100$in which there are no elements of order greater than$10$is the group$\mathbf{Z}_2\oplus\mathbf{Z}_2\oplus\mathbf{Z}_{5}\oplus\mathbf{Z}_{5}\cong \mathbf{Z}_{10}\oplus\mathbf{Z}_{10}$. Or I may be misunderstanding the seecond part, and instead you are told that$G$has no elements of order greater than$10$... the only possible orders, by Lagrange's Theorem, are$1$,$2$,$4$,$5$,$10$,$20$,$25$,$50$, and$100$. Since you are told there are no elements of order greater than$10$, then the orders must be$1$,$2$,$4$,$5$, or$10$. But if you have an element$x$of order$4$, then$x+b$is of order$20$(same argument as above), a contradiction. So every element is of order$1$,$2$,$5$, or$10$. And there are certainly elements of order$1$(namely,$0$), order$2$and$5$(Cauchy's Theorem), and order$10\$ (first part of the problem).

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You have systematically gone through each one of the steps I thought of in order, and in the time I took to think of the steps, you gave the answer. (+1) –  Mark Bennet May 10 '12 at 19:06