Subgroups, cosets, Lagrange, cyclic groups

In one sentence

Lagrange says the order of a subgroup divides the order of the group, the converse is false, and $A_4$ having no subgroup of order $6$ is the fact the whole topic is built to test.

Why the exam asks it

Because Lagrange is the first real theorem in group theory and its converse is the first real trap. "$d$ divides $|G|$, so there is a subgroup of order $d$" feels like a restatement and is not, and the counterexample has order $12$ — small enough to be asked about directly.

The idea

A subgroup $H \le G$ is a non-empty subset closed under the operation and inverses. For finite $H$, closure alone suffices.

The left cosets $gH$ partition $G$, and each has $|H|$ elements. Counting gives

$$|G| = [G:H]\,|H| \qquad \text{(Lagrange)},$$

where $[G:H]$ is the number of cosets. Two consequences used constantly:

• The order of any element divides $|G|$, since $\langle g\rangle$ is a subgroup.
• $g^{|G|} = e$ for every $g$. Applied to $(\mathbb{Z}/n)^\times$ this is Euler's theorem, and with $n$ prime it is Fermat's little theorem.

The converse fails

$|A_4| = 12$ and $6 \mid 12$, and $A_4$ has no subgroup of order 6.

The argument is worth carrying because it is short. A subgroup of order $6$ would have index $2$, hence be normal, hence contain every square. But every $3$-cycle in $A_4$ is a square — $(a\,b\,c) = (a\,c\,b)^2$ — and there are eight of them, which already exceeds $6$.

What is true in that direction:

• Cauchy: if a prime $p$ divides $|G|$, then $G$ has an element of order $p$. So subgroups of prime order always exist.
• Sylow: if $p^k$ is the largest power of $p$ dividing $|G|$, a subgroup of order $p^k$ exists.
• For cyclic groups the converse holds fully: $C_n$ has exactly one subgroup of order $d$ for every $d \mid n$, and no others.

So $6$ fails in $A_4$ specifically because $6$ is neither a prime nor a prime power, and $A_4$ is not cyclic.

Cyclic groups

$G = \langle g\rangle$ is cyclic. Every cyclic group is $\mathbb{Z}$ or $C_n \cong \mathbb{Z}/n$.

• Every subgroup of a cyclic group is cyclic.
• $C_n$ has exactly one subgroup of order $d$ for each $d \mid n$, generated by $g^{n/d}$.
• $g^k$ generates $C_n$ exactly when $\gcd(k,n) = 1$, so there are $\varphi(n)$ generators.
• The order of $g^k$ is $n/\gcd(k,n)$.

Every group of prime order is cyclic, by Lagrange: any non-identity element generates a subgroup of order dividing $p$ and bigger than $1$.

Small orders

• Order $p$: cyclic, one group.
• Order $p^2$: abelian, so $C_{p^2}$ or $C_p\times C_p$.
• Order $4$: $C_4$ or the Klein four-group $V$. Not every group of order $4$ is cyclic, and $V$ is the standard reminder.
• Order $6$: $C_6$ or $S_3$. The smallest non-abelian group is $S_3$.

Where intuition breaks

"$d$ divides $|G|$, so $G$ has a subgroup of order $d$." $A_4$ and $d = 6$. The single most examined false statement in algebra. Tempting because Lagrange reads like an equivalence and because it is one for cyclic groups.

"$G$ has order $n$ and $g^n = e$, so $g$ has order $n$." $g^n = e$ holds for every element by Lagrange. The order is the least such exponent.

"Every group of order 4 is cyclic." The Klein four-group. Every group of order $p^2$ is abelian, which is weaker than cyclic.

"$HK$ is a subgroup whenever $H$ and $K$ are." Only when $HK = KH$. In $S_3$, $\langle(1\,2)\rangle\langle(1\,3)\rangle$ has four elements, which does not divide $6$, so it cannot be a subgroup.

"A subgroup of index 2 might not be normal." Index $2$ always gives normality — the two left cosets and the two right cosets are both $\{H, G\setminus H\}$. Worth knowing as a true statement, since it appears among false ones.

"The intersection of two subgroups can fail to be a subgroup." Intersections are always subgroups; unions are the ones that fail, unless one contains the other.

"Cauchy gives a subgroup of order $d$ for every $d$." For every prime $d$ dividing $|G|$. Sylow extends it to prime powers, and nothing extends it to $6$.

The exam's angle

1. When an option asserts a subgroup of order $d$ exists, check whether $d$ is a prime power. If it is, Cauchy or Sylow gives it. If not, suspect $A_4$.
2. Check whether the group is cyclic first. In a cyclic group every divisor question has the obvious answer, and the paper often supplies $\mathbb{Z}/n$ to see whether you notice.
3. Use orders of elements to rule things out. A group with an element of order $n$ and size $n$ is cyclic; a group where every non-identity element has order $2$ is elementary abelian.
4. Count. Coset counting settles more of this topic than any structural argument, and the $A_4$ proof is itself a count.
5. For "how many generators", it is $\varphi(n)$, and for "how many subgroups of order $d$" in a cyclic group, exactly one.

The night before

• Lagrange: $|G| = [G:H]|H|$. Order of an element divides $|G|$; $g^{|G|} = e$.
• The converse is false: $A_4$ has order $12$ and no subgroup of order $6$.
• Cauchy: prime $p \mid |G|$ $\Rightarrow$ an element of order $p$. Sylow: a subgroup of order $p^k$ for the largest such power.
• Cyclic groups: converse of Lagrange holds, exactly one subgroup per divisor, all subgroups cyclic.
• $g^k$ generates $C_n$ $\iff \gcd(k,n)=1$; there are $\varphi(n)$ generators; $g^k$ has order $n/\gcd(k,n)$.
• Prime order $\Rightarrow$ cyclic. Order $p^2$ $\Rightarrow$ abelian, not necessarily cyclic.
• $HK$ is a subgroup $\iff HK = KH$. Intersections always are; unions almost never.
• Index $2$ $\Rightarrow$ normal.