# Finite abelian group

• Aug 22nd 2011, 04:42 AM
abhishekkgp
Finite abelian group
Let $\displaystyle A$ be a finite abelian group and let $\displaystyle p$ be a prime. Prove that the number of subgroups of $\displaystyle A$ of order $\displaystyle p$ equals the number of subgroups of $\displaystyle A$ of index $\displaystyle p$.
• Aug 22nd 2011, 01:10 PM
Deveno
Re: Finite abelian group
although this is not "quite" the same problem, i believe it is similar enough to help you: finite abelian group
• Aug 22nd 2011, 07:57 PM
ModusPonens
Re: Finite abelian group
If you have Hungerford's Algebra, look at the theorem II.2.6.(iii). Since A (or G in the book's notation) is finite, the F goes away. Now count the subgroups of index p and the subgroups of order p.
• Aug 22nd 2011, 10:42 PM
abhishekkgp
Re: Finite abelian group
Quote:

Originally Posted by Deveno
although this is not "quite" the same problem, i believe it is similar enough to help you: finite abelian group

In the link you have given it is first defined $\displaystyle A^p=\{a^p|a \in A\}$ and $\displaystyle A_p=\{a \in A| a^p=1\}$. Then it is claimed that $\displaystyle A_p \cap A^p =\{e \}$. I think there's a mistake here. Can you please check.
Also, I understand that $\displaystyle A/A_p \cong A^p$ but i can't see how it is useful in solving the question i had posted originally. Please help.
Thanks.
• Aug 24th 2011, 07:29 AM
abhishekkgp
Re: Finite abelian group
Quote:

Originally Posted by ModusPonens
If you have Hungerford's Algebra, look at the theorem II.2.6.(iii). Since A (or G in the book's notation) is finite, the F goes away. Now count the subgroups of index p and the subgroups of order p.

I can prove that $\displaystyle E_{p^n}$ has a total of $\displaystyle (p^n-1)/(p-1)$ distinct subgroups of order $\displaystyle p$.
To solve the original question(post #1) by the approach you have suggested i need to find the number of subgroups( of course distinct) of order $\displaystyle p$ of abelian groups like $\displaystyle Z_{p^3} \times Z_{p^2}$ etc. What i have found out (which i am not 100% sure is correct) that number of subgroups of order $\displaystyle p$ of $\displaystyle Z_{p^3} \times Z_{p^2}$ is same as the number of subgroups of order $\displaystyle p$ of $\displaystyle Z_p \times Z_p$ etc. is this correct??

Now i couldn't figure out how to find the number of subgroups having index $\displaystyle p$. Can you please help on this one??
• Aug 24th 2011, 05:49 PM
ModusPonens
Re: Finite abelian group
Quote:

Originally Posted by abhishekkgp
I can prove that $\displaystyle E_{p^n}$ has a total of $\displaystyle (p^n-1)/(p-1)$ distinct subgroups of order $\displaystyle p$.
To solve the original question(post #1) by the approach you have suggested i need to find the number of subgroups( of course distinct) of order $\displaystyle p$ of abelian groups like $\displaystyle Z_{p^3} \times Z_{p^2}$ etc. What i have found out (which i am not 100% sure is correct) that number of subgroups of order $\displaystyle p$ of $\displaystyle Z_{p^3} \times Z_{p^2}$ is same as the number of subgroups of order $\displaystyle p$ of $\displaystyle Z_p \times Z_p$ etc. is this correct??

Now i couldn't figure out how to find the number of subgroups having index $\displaystyle p$. Can you please help on this one??

I have to say that I thought I knew the answer to this problem when I posted the hint. But I realised later that I didn't. Anyway, I can say that $\displaystyle Z_{p^3} \times Z_{p^2}$ has more subgroups of order p than $\displaystyle Z_p \times Z_p$ because the latter has only $\displaystyle Z_p \times {0}$, $\displaystyle {0} \times Z_p$ and $\displaystyle <(1,1)>$, while the first has at least $\displaystyle <(p^2,p)>, <(p^2,0)>, <(0,p)>$ and $\displaystyle <(p^2,k \times p)>$ for k<p. Plus the analogous on the side of $\displaystyle p^2$

If I would have to resolve this problem, I would try to find, with specific examples, such as $\displaystyle Z_{5^2} \times Z_{5^3}$, the group of index 5 for each group of order 5 and thus understand the one to one correspondence.
• Aug 24th 2011, 07:12 PM
abhishekkgp
Re: Finite abelian group
Quote:

Originally Posted by ModusPonens
I have to say that I thought I knew the answer to this problem when I posted the hint. But I realised later that I didn't. Anyway, I can say that $\displaystyle Z_{p^3} \times Z_{p^2}$ has more subgroups of order p than $\displaystyle Z_p \times Z_p$ because the latter has only $\displaystyle Z_p \times {0}$, $\displaystyle {0} \times Z_p$ and $\displaystyle <(1,1)>$, while the first has at least $\displaystyle <(p^2,p)>, <(p^2,0)>, <(0,p)>$ and $\displaystyle <(p^2,k \times p)>$ for k<p. Plus the analogous on the side of $\displaystyle p^2$

If I would have to resolve this problem, I would try to find, with specific examples, such as $\displaystyle Z_{5^2} \times Z_{5^3}$, the group of index 5 for each group of order 5 and thus understand the one to one correspondence.

nooooo. $\displaystyle Z_p \times Z_p$ has exactly $\displaystyle p+1$ subgroups of order $\displaystyle p$. namely $\displaystyle \left<(x,0) \right>, \left<(x,y)\right>, \left<(x,y^2)\right>, \ldots, \left<(x,y^{p-1})\right>, \left<(0,y)\right>$
• Aug 25th 2011, 05:29 PM
ModusPonens
Re: Finite abelian group
You're right. Sorry.

I don't have the time necessary to devote myself to the problem, so I will leave it as it is.
• Aug 25th 2011, 05:48 PM
Drexel28
Re: Finite abelian group
Quote:

Originally Posted by ModusPonens
You're right. Sorry.

I don't have the time necessary to devote myself to the problem, so I will leave it as it is.

If interested, I provided the necessary tools to solve this in this thread.