# Supose that G is a finite abelian group that does not contain a subgro

1. Apr 20, 2014

### vish_maths

let us assume G is not cyclic. Let a be an element of G of maximal order. Since G is not cyclic we have <a>≠G. Let b be an element in G, but not in the cyclic subgroup generated by a.

O(a) = m and O(b) = n where O() refers tothe orders. . then how can we use this to construct a subgroup of G isomorphic to Zp×Zp?

Help will be appreciated. Thank you

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2. Apr 20, 2014

### Dick

Can you prove that if the prime factorization of |G| is $p_1 p_2 p_3...p_n$ and all of the primes $p_i$ are different, then G is cyclic? If two are the same, then what?

Last edited: Apr 20, 2014
3. Apr 20, 2014

### vish_maths

If |G| =p1p2....pn and all of the primes are different, then |H| will be one of those primes. Since a group of prime order is cyclic, then H will be cyclic in that case . Infact all proper subgroups of G will be cyclic.
but i am not sure how to prove G as cyclic with this data..

4. Apr 21, 2014

### micromass

Staff Emeritus
What does the fundamental theorem of Abelian groups tell you?

5. Apr 21, 2014

### vish_maths

the book which i am reading ( GAllian ) has not introduced this topic as of yet. In fact, not even normal and factor groups. i am on the chapter on external direct products.

6. Apr 21, 2014

### micromass

Staff Emeritus
Where exactly in Gallian is this?

7. Apr 21, 2014

### vish_maths

Supplementary exercise for chapters 5 - 8 . Question no. 50 . Gallian 7/e contemporary guide to abstract algebra

8. Apr 21, 2014

### vish_maths

and pg. no 50

9. Apr 21, 2014

### micromass

Staff Emeritus
OK. So let $G$ be finite abelian group such that $G$ does not contain a subgroup isomorphic to $\mathbb{Z}_p\times \mathbb{Z}_p$ for any prime. Consider the prime factorization $|G|= p_1^{a_1} ... p_k^{a_k}:= p_1^{a_1}m$.

Define $H = \{x\in G~\vert~\text{order of}~H~\text{divides}~p_1^{a_1}\}$ and $K = \{x\in G~\vert~\text{order of}~K~\text{divides}~m\}$.

Start by proving the following:
1) $H$ and $K$ are subgroups of $G$.
2) For any $x\in G$ and integers $s$ and $t$, we have $x^{sp^k}\in H$ and $x^{tm}\in K$. Deduce that $G=HK$.
3) Show that $H\cap K = \{e\}$.
4) Find $|H|$ and $|K|$.
5) Prove that $H$ is cyclic.
6) Prove that $G$ is the product of cyclic groups
7) Prove the result.

Last edited: Apr 21, 2014
10. Apr 21, 2014

### Dick

Your definition of K doesn't make much sense. I assume you mean $K = \{x\in G~\vert~\text{order of}~x~\text{divides}~m\}$.

11. Apr 21, 2014

### micromass

Staff Emeritus
Yes, sorry!