10

I've just started learning about groups in abstract algebra and today we were discussing subgroups. One of the examples we used in class was $(\Bbb Z_6,+)$, or integers mod six under addition. In addition to the trivial and improper subgroups, we found two additional subgroups:

$$\{0,1,2,3,4,5\}$$

$$\{0,2,4\}$$

$$\{0,3\}$$

$$\{0\}$$

What I found particularly interesting is that when looking at the cardinality of all four subgroups, they directly corresponded to the four factors of six. I tried thinking of other examples and each time it was the same story. What I want to know is if this generally holds for all $\Bbb Z_n\;\;(n\in\Bbb N)$, and if so, why it holds and what this means.

amWhy
  • 210,739
Leonidas Lanier
  • 4,296
  • 6
    +1 for the observation. If you continue to pay this kind of attention to the theorems (and, particularly, the examples) in your abstract algebra course you may well find yourself guessing/anticipating theorems you will see in the weeks to come. In particular, you'll soon get to Lagrange's theorem. as some of the answers point out. You should be pleased, not discouraged, when answers like that refer to stuff you haven't yet studied. – Ethan Bolker Apr 14 '17 at 18:07

3 Answers3

8

It's true for all finite groups that every subgroup has order dividing that of the main group. This is called Lagrange's theorem. The converse is not true of all finite groups.

A finite group of size $n$ which has subgroups of size $d$ for all $d \mid n$ is called a Lagrangian group or a CLT group ("Converse to Lagrange's Theorem"). They're still not completely characterised, but it's known that they're all solvable.

However, two general results tell us some things about subgroups which have certain sizes:

  • Cauchy's theorem tells us that whenever $p \mid n$, there is a subgroup of size $p$;
  • Sylow's theorems tell us that whenever $p^i \mid n$ is the highest power of $p$ dividing $n$, then there is a subgroup of size $p^i$.

All finite abelian groups are Lagrangian (including $\mathbb{Z}_n$ for all $n$, where the proof is simple: just take the subgroup generated by $\frac{n}{d}$, which has order $d$).

Related SE question which contains much more detail: Complete classification of the groups for which converse of Lagrange's Theorem holds.

I'd like to call attention to Ethan Bolker's comment on your question, which I wholeheartedly endorse:

+1 for the observation. If you continue to pay this kind of attention to the theorems (and, particularly, the examples) in your abstract algebra course you may well find yourself guessing/anticipating theorems you will see in the weeks to come. In particular, you'll soon get to Lagrange's theorem. as some of the answers point out. You should be pleased, not discouraged, when answers like that refer to stuff you haven't yet studied.

Patrick Stevens
  • 36,855
4

Notice that the generator for your subgroup of size $3$ was $2$, size $1$ was generated by $6 \equiv 0$, size $6$ generated by $1$, and size $2$ generated by $3$. This will hold in general - if $f$ is a factor of $n$ (ie, $f | n$), then $\frac{n}f$ is an integer, and hence generates an additive subgroup of $\mathbb{Z}_n$. This will have order $f$, since $\frac{n}f f = n$, and for $0 \leq c < f$, $c \frac{n} f$ will give you some value less than $n$. Hence the subgroup has order $f$.

Artimis Fowl
  • 1,960
4

This is a great example of Lagrange's Theorem! Simply stated, the theorem is

For a finite group $G$ and a subgroup $H<G$, the order of $H$ divides the order of $G$.

Now, why should this be? The main idea is that you can construct an equivalence relation $\sim$ on $G$ where $a\sim b$ if and only if $a=bh$ for some $h\in H$.

Try to prove Lagrange's theorem using this relation!

Santana Afton
  • 6,978