Find the number of subgroups of $\mathbb{Z}_p\oplus \mathbb{Z}_p \oplus\mathbb{Z}_p$ ($p$ is a prime number). How about the number of subgroups up to isomorphism? and what the most generalized case? My idea is to prove $(a,b,c)$ can generate $p^3$ when $a,b,c$ are not equal to $0$, then we discuss the cases when one of them or two of them are $0$. but I still do not know how to generalize it. for any group $G$ whose order is $n$, can we find how many subgroup it has? Thank you for all your help.
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What have you tried? This sounds like a homework problem, so please edit the question with your thoughts concerning how to approach the problem, and specific questions as to where you are getting stuck, so we can further assist you in understanding the subject. – Nicholas Stull May 09 '14 at 04:19
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Here's another way of solving this problem. Notice that $\newcommand{\Zp}{\mathbb{Z}_p} \Zp \oplus \Zp \oplus \Zp$ is a vector space over the field $\Zp$. We are looking for the number of subspaces of this vector space. The vector space has dimension $3$.
The number of subspaces of this vector space of dimension $k$ is given by the number of linearly independent lists of vectors of length $k$, divided by the number of linearly independent lists of size $k$ than can be found within any particular subspace.
The number of linearly independent sets of size $k$ is given by $$ (p^3 - 1)(p^3 - p)(p^3 - p^2)\cdots (p^3 - p^{k-1}). $$
The number of linearly independent sets of size $k$ within a given subspace of dimension $k$ is given by $$ (p^k - 1)(p^k - p)(p^k - p^2) \cdots (p^k - p^{k-1}). $$
In our case, we are interested in the cases $k = 0, k = 1, k = 2, k = 3$. So the total number of subspaces is \begin{align*} \sum_{k=0}^3 &\frac{(p^3 - 1)(p^3 - p)(p^3 - p^2)\cdots (p^3 - p^{k-1})} {(p^k - 1)(p^k - p)(p^k - p^2) \cdots (p^k - p^{k-1})} \\ &= 1 + \frac{p^3 - 1}{p - 1} + \frac{(p^3 - 1)(p^3 - p)}{(p^2 - 1)(p^2 - p)} + 1 \\ &= 1 + (p^2 + p + 1) + (p^2 + p + 1) + 1 \\ &= 2p^2 + 2p + 4. \end{align*}
See also this answer.
Caleb Stanford
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could you give me a hint how to use this method to solve any given order n. – Danial Jeo May 13 '14 at 04:33
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1@DanialJeo You mean where the original group is $p^n$ multiplied copies of $n$? Just replace $3$ with $n$. The expression you will get can be found in the other answer. – Caleb Stanford May 13 '14 at 05:09
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I mean,$n=p\bigoplus q \bigoplus \dots$, for any given n, n can factor into products of subgroups.@Goos – Danial Jeo May 13 '14 at 05:28
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@DanialJeo I am still not quite sure what you are asking, but it sounds like you are asking a new question. If you have a new question, ask a new question, and link back to this one. Regards – Caleb Stanford May 13 '14 at 05:31
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Any subgroup of $\newcommand{\Zp}{\mathbb{Z}_p} \Zp \oplus \Zp \oplus \Zp$ must have order $1, p, p^2,$ or $p^3$.
Order $\boldsymbol{1}$, Order $\boldsymbol{p^3}$ : One trivial subgroup in each case.
Order $\boldsymbol{p}$ : Any nonzero element generates a subgroup of order $p$, and since any subgroup of order $p$ has a nonzero element, all subgroups are generated by some element. Moreover a given subgroup of order $p$ is isomorphic to $\Zp$ and is generated by any nonzero element in the subgroup. Thus the total number of subgroups in this case is the total number of nonzero elements, divided by $p-1$ since each is counted once for each nonzero element.
Order $\boldsymbol{p^2}$ : There are no elements of order $p^2$, hence no subgroups isomorphic to $\mathbb{Z}_{p^2}$. So we are looking for subgroups isomorphic to $\Zp \oplus \Zp$. Try counting such groups by picking any element of the entire group, then picking an element not in the subgroup generated by that group. Then try to find the number of times each subgroup was counted in this way, and divide by this number.
Caleb Stanford
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