Finite abelian group

**abhishekkgp** · August 22nd 2011, 04:42 AM

Let $A$ be a finite abelian group and let $p$ be a prime. Prove that the number of subgroups of $A$ of order $p$ equals the number of subgroups of $A$ of index $p$ .

**Deveno** · August 22nd 2011, 01:10 PM

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

**ModusPonens** · August 22nd 2011, 07:57 PM

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.

**abhishekkgp** · August 22nd 2011, 10:42 PM

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 $A^p=\{a^p|a \in A\}$ and $A_p=\{a \in A| a^p=1\}$ . Then it is claimed that $A_p \cap A^p =\{e \}$ . I think there's a mistake here. Can you please check.
Also, I understand that $A/A_p \cong A^p$ but i can't see how it is useful in solving the question i had posted originally. Please help.
Thanks.

**abhishekkgp** · August 24th 2011, 07:29 AM

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 $E_{p^n}$ has a total of $(p^n-1)/(p-1)$ distinct subgroups of order $p$ .
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 $p$ of abelian groups like $Z_{p^3} \times Z_{p^2}$ etc. What i have found out (which i am not 100% sure is correct) that number of subgroups of order $p$ of $Z_{p^3} \times Z_{p^2}$ is same as the number of subgroups of order $p$ of $Z_p \times Z_p$ etc. is this correct??

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

**ModusPonens** · August 24th 2011, 05:49 PM

Originally Posted by abhishekkgp

I can prove that $E_{p^n}$ has a total of $(p^n-1)/(p-1)$ distinct subgroups of order $p$ .
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 $p$ of abelian groups like $Z_{p^3} \times Z_{p^2}$ etc. What i have found out (which i am not 100% sure is correct) that number of subgroups of order $p$ of $Z_{p^3} \times Z_{p^2}$ is same as the number of subgroups of order $p$ of $Z_p \times Z_p$ etc. is this correct??

Now i couldn't figure out how to find the number of subgroups having index $p$ . 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 $Z_{p^3} \times Z_{p^2}$ has more subgroups of order p than $Z_p \times Z_p$ because the latter has only $Z_p \times {0}$ , ${0} \times Z_p$ and $<(1,1)>$ , while the first has at least $<(p^2,p)>, <(p^2,0)>, <(0,p)>$ and $<(p^2,k \times p)>$ for k<p. Plus the analogous on the side of $p^2$

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

**abhishekkgp** · August 24th 2011, 07:12 PM

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 $Z_{p^3} \times Z_{p^2}$ has more subgroups of order p than $Z_p \times Z_p$ because the latter has only $Z_p \times {0}$ , ${0} \times Z_p$ and $<(1,1)>$ , while the first has at least $<(p^2,p)>, <(p^2,0)>, <(0,p)>$ and $<(p^2,k \times p)>$ for k<p. Plus the analogous on the side of $p^2$

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

nooooo. $Z_p \times Z_p$ has exactly $p+1$ subgroups of order $p$ . namely $\left<(x,0) \right>, \left<(x,y)\right>, \left<(x,y^2)\right>, \ldots, \left<(x,y^{p-1})\right>, \left<(0,y)\right>$

**ModusPonens** · August 25th 2011, 05:29 PM

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.

**Drexel28** · August 25th 2011, 05:48 PM

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.

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