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It is given that the order of some finite abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10.

It is clear that since order of group is divisible by 10. By converse to Lagrange's Theorem, if 10 divides the order of the group G, then G has a subgroup of order 10.

But to ensure that this subgroup is cyclic.

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"If $m$ divides the order of the group $G$, then $G$ has a subgroup of order $10$". Setting aside the fact that the converse of Lagrange's theorem does not hold in general (though it does hold for abelian groups, but it has to be proven for abelian groups), that sentence does not make much sense: the consequent is independent of $m$. –  Arturo Magidin Oct 10 '11 at 19:30
    
@Arturo Magidin: Thanks, for your reply. I have edited the question now. It is mentioned here that order of G is divisible by 10 and G is a finite abelian group, then using Lagrange's Theorem, we can infer that G has a subgroup H of order 10. Then, by Fundamental Theorem of Finite abelian groups, we can conclude that as H is a finite ableian, it is isomorphic to Z2 X Z5 ( external direct product), and is cyclic. –  Tav Oct 10 '11 at 19:48
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No: Lagrange's Theorem goes in the opposite direction. Lagrange's Theorem says that if $G$ is a finite group, and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$. You are trying to invoke Lagrange's Theorem to claim that if $m$ divides $|G|$, then $G$ has a subgroup of order $m$. That is not Lagrange's Theorem, and while the result does hold for abelian groups, the proof requires some work and it does not follow from Lagrange's Theorem. Have you actually proven this converse for abelian groups? –  Arturo Magidin Oct 10 '11 at 19:52
    
@ArturoMagidin: Thanks , you are right. I was trying to use the converse to the Lagrange's Theorem, which I guess holds for finite cyclic groups. I think a Corollary to Fundamental Theorem of Finite Abelian groups, which says that " If m divides the order of a finite abelian group G, then G has a subgroup of order m" can be used. Please suggest how to move ahead. –  Tav Oct 10 '11 at 20:05
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If you have that converse already proven, then you can go ahead (it can also be proven inductively by applying Cauchy's Theorem to construct a subgroup one prime at a time, so you don't need the full force of the Fundamental Theorem). Once you know $G$ has a subgroup of order $10$, you need to show an abelian group of order 10 must be cyclic. HINT: if $a$ has order $2$ and $b$ has order $5$ (such elements exist by Cauchy's Theorem), what is the order of $ab$? –  Arturo Magidin Oct 10 '11 at 20:21
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3 Answers

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Hint:If $p$ and $q$ are distinct primes then $\mathbb{Z}_p\times\mathbb{Z}_q\cong \mathbb{Z}_{pq}$.

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$G$ has a subgroup of order $2$ and order $5$ by Cauchy's theorem. Since $2$ is prime to $5$, the order of the product of two generators of these groups is $10$ and it generates then a (cyclic) subgroup of $G$ of order $10$.

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Lagrange's theorem doesn't guarantee the existence of subgroups of order 2 and 5 (the converse of Lagrange's theorem isn't true in general). Cauchy's theorem does however give us elements of orders 2 and 5, and thus subgroups of orders 2 and 5. –  SL2 Oct 10 '11 at 19:27
    
k, just didn't remember whose that theorem was from... :p changed the name of the result. Thanks. –  Louis La Brocante Oct 10 '11 at 19:29
    
Note that, to conclude that the product of an element of order 2 and an element of order 5 has order 10, it is very important that the group be abelian. Otherwise the order of the product could be anything. –  Greg Martin Oct 10 '11 at 20:10
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Hint: What abelian groups of order 10 can you think of?

(There is only one, up to isomorphism.)

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Thanks for your reply. By Fundamental Theorem of finite abelian group, there is only one Z2+Z5 of order 10, which is cyclic. –  Tav Oct 10 '11 at 19:54
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@Tav: Precisely! –  Zev Chonoles Oct 10 '11 at 19:54
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