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If $G$ is an abelian group of order 72, do we know how many subgroups of order 8 it has?

Just because it's a divisor doesn't mean that there is a subgroup of that size. But I'm wrong. Why?

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Take a look at the Sylow theorems. – Harald Hanche-Olsen Oct 2 '12 at 10:23

In general, it is not true that the converse of Lagrange's theorem holds, ie that for any for any divisor $d$ of the order of the finite group $G$, $G$ has a subgroup of order $d$.

However, if $G$ is abelian, the converse of Lagrange's theorem does hold. Here is an elementary proof:

First, we want to show Cauchy's theorem for abelian groups, ie that if $p$ is a prime dividing the order of $G$ the $G$ has a subgroup of order $p$. In order to do this, we assume for the sake of contradiction that it does not hold and let $G$ be a minimal counter example. This means that $p$ divides $|G|$ and $G$ has no subgroup of order $p$, but if $H$ is a subgroup of $G$ and $p$ divides $|H|$, then $H$ has a subgroup of order $p$. We immediately see that this means that for any proper subgroup $H$ of $G$, we have that $p$ does not divide $|H|$. In order to obtain the desired contradiction, let $H$ and $K$ be maximal proper subgroups of $G$ with $H\neq K$ (see the end of the proof for an explanation of why we can assume $G$ has such subgroups). Now $p$ divides neither $|H|$ nor $|K|$, but since $G$ is abelian, we know that $HK$ is a subgroup of $G$ of order $\frac{|H||K|}{|H\cap K|}$, and since $H$ and $K$ are maximal and not equal, this subgroup must be $G$ itself. But clearly $\frac{|H||K|}{|H\cap K|}$ is not divisible by $p$ since neither of $|H|$ and $|K|$ are, which contradicts our assumption that $|G|$ is divisible by $p$.

In the proof, I claimed that we could assume that $G$ had two distinct maximal proper subgroups. To see this, let us deal with the case where $G$ has exactly one maximal proper subgroup $H$. I claim that in this case $G$ is cyclic, because if we look at any element $g\in G$ which is not in $H$, then $\left< g\right>$ is a cyclic subgroup of $G$, and if it was a proper subgroup, it would be contained in a maximal proper subgroup. But the unique maximal proper subgroup of $G$ is $H$, and since $g$ is not in $H$, this cannot be the case. Hence $\left< g\right>$ is $G$, which is then cyclic.

So we need to show that if $G$ is a cyclic group and $p$ divides the order of $G$, then $G$ has a subgroup of order $p$, which can be done as follows: Let $g$ be a generator of $G$ and let $n = \frac{|G|}{p}$. Since $g$ is a generator, $g^n$ is not the identity. But $(g^n)^p = g^{|G|}$ is the identity, so we have produced an element of order $p$, which then generates a subgroup of order $p$ as desired.

We can now use thew above to prove that if $n$ divides $|G|$ then $G$ has a subgroup of order $n$, by induction. Let $p$ be a prime divisor of $n$ and let $H$ be a subgroup of $G$ of order $p$ (which exists by the above). Now $G/H$ is a group of strictly smaller order than $G$, and it has order divisible by $\frac{n}{p}$, so by induction it has a subgroup of order $\frac{n}{p}$. This subgroup then corresponds to a subgroup of $G$ be the usual isomorphism theorems, and the subgroup of $G$ it corresponds to has order $p\frac{n}{p} = n$ as desired.

In order to determine the number of subgroups of a given order in an abelian group, one needs to know more than the order of the group, since for example there are two different groups of order $4$, and one of them has one subgroup of order $2$, which the other has $3$.

However, in certain cases, it is possible to determine it just from the order of the group, as is the case in the question. Here, the order sought is a number $n$ such that $n$ and $\frac{|G|}{n}$ are coprime, and in this case it is easy to check that $G$ has exactly one subgroup of order $n$, for example by using the classification of finite abelian groups (or the generalizations of the Sylow theorems due to Hall, but these require a lot more and are usually not taught in introductory courses)

Edit: I realized that we can actually avoid using anything near as powerful as the classification for the above. Using only elementary means, we can show the following, which has a couple of nice immediate corollaries:

Let $G$ be a finite group and $H$ a normal subgroup of $G$ such that $|H|$ and $\frac{|G|}{|H|}$ are coprime. Let $K$ be a subgroup of $G$ such that $|K|$ divides $|H|$. Then $K\subseteq H$.

Proof: We have that $|HK| = \frac{|H||K|}{|H\cap K|}$ and since $H$ is a normal subgroup of $G$, $HK$ is also a subgroup of $G$. But now the order of $HK$ is $k|H|$ where $k = \frac{|K|}{|H\cap K|}$ is a divisor of $|H|$, which means that we must have $k = 1$ as this is the only divisor of $|H|$ that also divides $\frac{|G|}{|H|}$, and this exactly means that $|H\cap K| = |K|$, ie that $K\subseteq H$.

Now as mentioned, we get that if $G$ is abelian, and $H$ is a subgroup of $G$ whose order is coprime to its index, then $G$ has no other subgroup of the same order as $H$.

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Denote $C_n$ the cyclic group of order $n$. It is unique up to isomorphism.

According to the structure theorem for finite groups, the following are the only possibilities for an abelian group with $72$ elements:

  • $G\simeq C_{72}$,
  • $G\simeq C_{36}\times C_2$,
  • $G\simeq C_{18}\times C_2\times C_2$,
  • $G\simeq C_{24}\times C_3$,
  • $G\simeq C_{12}\times C_6$,
  • $G\simeq C_{6}\times C_6\times C_2$, .

Since $C_n$ has only one subgroup with $d$ elements for every divisor $d$ of $n$, it is now very easy to compute the number of subgroups of $G$ with $8$ elements in every situation.

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Sry ... but there have to be six ones, right? So $G \cong C_6 \times C_3 \times C_2 \times C_2$ is missing. – martini Oct 2 '12 at 11:43
No! $3$ divides $6$ but $2$ does not divide $3$ ! – Andrea Mori Oct 2 '12 at 11:47
but you're still correct in the sense that $C_3\times C_2\simeq C_6$ and that's the sixth case that I had missed. Thanks! – Andrea Mori Oct 2 '12 at 11:51

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