Suppose you have an abelian group of size 1,000. How many subgroups does it have?

I know there are 9 such groups from $1,000 = 2^3 \times 5^3$ giving us 3 of order $2^3 \times$ 3 of order $5^3$ because there are 3 integer partitions of 3 but I'm not sure if that is helpful.

From the Fundamental Theorem of Finitely Generated Abelian groups. I know that there are cyclic subgroups $\mathbb{Z}_8, \mathbb{Z}_{125}$. I think we also have $\mathbb{Z}_2 \times \mathbb{Z}_4$ because 2 and 4 are not co-prime so it's not isomorphic to $\mathbb{Z}_8$ but I'm not sure of the equivalent group of order 125.

I don't know what other non-cyclic subgroups it might have

  • 2
    $\begingroup$ You should be aware of the fact that there exist many non-isomorphic abelian groups of order $1000$. Begin with the cyclic one: there is a complete description of the subgroups of a cyclic group. $\endgroup$ – Crostul Jun 17 '15 at 7:37
  • $\begingroup$ Is the original group cyclic? $\endgroup$ – Mark Bennet Jun 17 '15 at 7:38
  • 1
    $\begingroup$ We were only given that it was abelian. Is the cyclic group of order 1,000 the one generated by $\mathbb{Z}_8 \times \mathbb{Z}_{125}$? $\endgroup$ – Alyx Captain Jun 17 '15 at 7:41
  • $\begingroup$ This question has been extensively discussed in more generality, see here. $\endgroup$ – Dietrich Burde Jun 17 '15 at 7:58

Assume $A$ is abelian of order $1000$ then we have :

$$A=S_2\times S_5 $$

where $S_2$ and $S_5$ are respectively the $2$ and $5$-Sylow subgroup. Now I claim that any subgroup $H$ will be written as :

$$H=H_2\times H_5 \text{ where } H_p:=H\cap S_p\text{ for } p=2,5$$

(there is something to prove here, I leave this to you).

Hence we see that there is canonical bijection between :

$$\{\text{subgroups of }G\}\text{ and }\{\text{subgroups of }S_2\}\times \{\text{subgroups of }S_5\}$$

Hence it reduces to understand the number of subgroups (for $p$ prime) of the three different abelian groups of order $p^3$ :

$$\mathbb{Z}_{p^3}\text{, }\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}\text{ and } \mathbb{Z}_{p}\times \mathbb{Z}_{p}\times \mathbb{Z}_{p} $$

The first one has exactly four subgroups (hint : exactly one for each divisor). The third one has $1+\frac{p^3-1}{p-1}+\frac{p^3-1}{p-1}+1=2p^2+2p+4$ subgroups (hint : a subgroup of the third group is a $\mathbb{Z}_{p}$-subvectorial space). The second one, it is a little harder.

Set $G=\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$. Take $H$ a subgroup of $G$, the exposant of $H$ is either $1$, $p$, $p^2$. There is exactly one subgroup of exposant $1$ (the trivial one).

Now if $H$ is a subgroup of exposant $p$ then $H$ must be included in :

$$p\mathbb{Z}_{p^2}\times \mathbb{Z}_p $$

This last group is isomorphic to $\mathbb{Z}_{p}\times \mathbb{Z}_p$ and has exactly $\frac{p^2-1}{p-1}+1=p+2$ subgroups of exposant $p$. Hence we have in $G$ $p+2$ subgroups of exposant $p$.

Now if $H$ is a subgroup of exposant $p^2$ it is either of order $p^3$ (in which case $H=G$) or is of order $p^2$ hence cyclic. It is somehow easy to show that $H$ will always contain a unique element of the form $(1,a)$ and that such an element is of order $p^2$. Hence we have as many cyclic subgroup of order $p^2$ in $G$ as the number of choice for $a$, i.e. $p$ ($a\in \mathbb{Z}_p$). Finally we have $p+1$ subgroup of exposant $p^2$. Finally we have $2p+4$ subgroups in $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$.

To sum up

$$\mathbb{Z}_{p^3}\text{ has }4\text{ subgroups}$$

$$\mathbb{Z}_{p}\times \mathbb{Z}_{p}\times \mathbb{Z}_{p}\text{ has } 2p^2+2p+4\text{ subgroups}$$

$$\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}\text{ has } 2p+4\text{ subgroups}$$

In particular if you know your abelian group of order $1000$ (e.g. its decomposition through the theorem of finitely generated abelian groups) you can easily deduce the form of the $p$-Sylow and apply the two results above. For instance take :

$$ G:=\mathbb{Z}_5\times\mathbb{Z}_{10}\times\mathbb{Z}_{20}$$

Then we have that :

$$G=(\mathbb{Z}_2\times\mathbb{Z}_4)\times(\mathbb{Z}_5\times\mathbb{Z}_5\times\mathbb{Z}_5) $$

Hence we have exactly $((2\times 2+4)+(2\times 5^2+2\times 5+4))$ subgroups, i.e. $8+50+10+4=72$ subgroups.


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