abelian group

**poorna** · May 25th 2009, 06:17 AM

Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.

**Swlabr** · May 25th 2009, 06:46 AM

Use the fundamental theorem for finite Abelian groups twice. Once to show that it is sufficient to prove the result for $p$ -groups, and the second time to investigate said $p$ -groups.

Note also that if $p$ divides the order of a group then that group contains an element of order $p$ .

You would have been better posting this in the abstract algebra forum, btw...

**NonCommAlg** · May 25th 2009, 09:46 AM

Originally Posted by poorna

Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.

let $n=\max \{o(g): \ g \in G \}$ and $o(a)=n.$ if $n=|G|,$ then we're done. otherwise, choose $b \notin <a>.$ let $o(b)=m.$ if $m \mid n,$ then $o(b)=o(a^{n/m})$ and thus by the uniqueness property of G

we must have $b \in <a^{n/m}>,$ which is not possible. so $m \nmid n$ and hence $\text{lcm}(n,m) > n.$ but, since G is abelian, there exists an element of G of order ....................... contradiction!

**TheAbstractionist** · May 25th 2009, 03:30 PM

Another proof?

Let $|G|=n.$ If $G$ is not cyclic, then $G$ would be the internal direct product of distinct cyclic subgroups $C_{n_1}C_{n_2}\cdots C_{n_r}$ where $n_i\mid n_{i-1}$ and $n=n_1n_2\cdots n_r.$ Since $n_2\mid n_1,$ it follows that $C_{n_1}$ would have a subgroup of order $n_2$ and so $G$ would have two subgroups of order $n_2.$ Contradiction!

**NonCommAlg** · May 25th 2009, 11:11 PM

Originally Posted by TheAbstractionist

Another proof?

Let $|G|=n.$ If $G$ is not cyclic, then $G$ would be the internal direct product of distinct cyclic subgroups $C_{n_1}C_{n_2}\cdots C_{n_r}$ where $n_i\mid n_{i-1}$ and $n=n_1n_2\cdots n_r.$ Since $n_2\mid n_1,$ it follows that $C_{n_1}$ would have a subgroup of order $n_2$ and so $G$ would have two subgroups of order $n_2.$ Contradiction!

yes, but i was actually avoiding using the fundamental theorem of finite(ly generated) abelian groups.

now, i was thinking, what if we try to make poorna's question more interesting:

this time suppose $G$ is a finitely generated abelian group such that for any positive integer $n$ there exists at most one subgroup $H$ of $G$ with $[G:H]=n.$ what can we say about $G$ ?

**Swlabr** · May 26th 2009, 03:18 AM

Originally Posted by NonCommAlg

this time suppose $G$ is a finitely generated abelian group such that for any positive integer $n$ there exists at most one subgroup $H$ of $G$ with $[G:H]=n.$ what can we say about $G$ ?

Then $G = I(G) \times C_m, m \geq 0 \in \mathbb{N}$ where $I(G)$ is the subgroup of all non-torsion elements (along with the identity). $I(G)$ is a subgroup because $G$ is abelian. We get that $G = I(G) \times C_m$ by inducting on the number of generators of the group, $n$ . Clearly the result holds for a 1-generated group.

Assume the result holds for all groups generated by less than or equal to $n$ elements. Let $G$ be a subgroup generated by $n+1$ elements. Then, $I(G) \unlhd G$ is of finite index as $G/I(G)$ contains only torsion elements. Thus, $G/I(G)$ is generated by less than or equal to $n$ elements (this can be seen by viewing $G$ as an $n+1$ -dimensional vector space). Is it of the form we wish? Let $[G/I(G):K_1] = [G/I(G):K_2] \leq \infty$ . Then as we are in an Abelian group these subgroup as normal and so we can apply the correspondence theorem to get that they are normal subgroups of $G$ : $K_1, K_2 \lhd G$ . Then, $[G:K_1] = [G:I(G)].[I(G):K_1] = [G:I(G)].[I(G):K_2] = [G:K_2]$ , and so they are the same subgroups. Thus, $G/I(G)$ is of the correct form. As it consists only torsion elements it is cyclic.

Viewing our group as a vector space we can glue these two bits together - their intersection is trivial and $G/I(G) + I(G) = G$ and so we have a direct sum, and so a direct product. Thus, $G = I(G) \times C_m$ .

I think we cannot do anything with $I(G)$ . Clearly it is just $C_{\infty} \times ... \times C_{\infty}$ , but $C_{\infty} \times ... \times C_{\infty}$ conforms to the conditions stipulated.

**NonCommAlg** · May 26th 2009, 11:59 AM

Originally Posted by Swlabr

Then $G = I(G) \times C_m, m \geq 0 \in \mathbb{N}$ where $I(G)$ is the subgroup of all non-torsion elements (along with the identity). $I(G)$ is a subgroup because $G$ is abelian. We get that $G = I(G) \times C_m$ by inducting on the number of generators of the group, $n$ . Clearly the result holds for a 1-generated group.

Assume the result holds for all groups generated by less than or equal to $n$ elements. Let $G$ be a subgroup generated by $n+1$ elements. Then, $I(G) \unlhd G$ is of finite index as $G/I(G)$ contains only torsion elements. Thus, $G/I(G)$ is generated by less than or equal to $n$ elements (this can be seen by viewing $G$ as an $n+1$ -dimensional vector space). Is it of the form we wish? Let $[G/I(G):K_1] = [G/I(G):K_2] \leq \infty$ . Then as we are in an Abelian group these subgroup as normal and so we can apply the correspondence theorem to get that they are normal subgroups of $G$ : $K_1, K_2 \lhd G$ . Then, $[G:K_1] = [G:I(G)].[I(G):K_1] = [G:I(G)].[I(G):K_2] = [G:K_2]$ , and so they are the same subgroups. Thus, $G/I(G)$ is of the correct form. As it consists only torsion elements it is cyclic.

Viewing our group as a vector space we can glue these two bits together - their intersection is trivial and $G/I(G) + I(G) = G$ and so we have a direct sum, and so a direct product. Thus, $G = I(G) \times C_m$ .

I think we cannot do anything with $I(G)$ . Clearly it is just $C_{\infty} \times ... \times C_{\infty}$ , but $C_{\infty} \times ... \times C_{\infty}$ conforms to the conditions stipulated.

yes, we have $G=T(G) \times \mathbb{Z}^m,$ where $T(G)$ is a a product of finite cyclic groups. then $G/\mathbb{Z}^m \cong T(G)$ has the same property as $G$ because $[G/\mathbb{Z}^m : H/\mathbb{Z}^m]=[G:H].$ thus $T(G)$ has to be

cyclic because it's finite. so $G=C_n \times \mathbb{Z}^m.$ now the question is: does any $G$ of this form have the property that if $[G:H_1]=[G:H_2] < \infty,$ for some subgroups $H_1,H_2$ of $G,$ then $H_1=H_2$ ?

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