Why Must a Group of Order 27 Have a Subgroup of Order 3?

Every group with 27 elements has a subgroup with exactly 3 elements. The quickest proof uses Lagrange’s theorem: the order of any element divides the order of the group. Since 27 is a power of 3, any nonidentity element has order 3, 9, or 27, and each possibility gives an element of order 3.

The resulting subgroup is cyclic. If $a$ is an element of order 3, then $\langle a\rangle=\{e,a,a^2\}$, where $e$ is the identity.

The proof using Lagrange’s theorem

Let $G$ be a group with $|G|=27$. Because the identity is only one of its elements, there is a nonidentity element $g\in G$. Write $|g|$ for the order of $g$, meaning the smallest positive integer $n$ for which $g^n=e$.

The subgroup $\langle g\rangle$ has $|g|$ elements. By Lagrange’s theorem, the order of a subgroup divides the order of the finite group containing it. Therefore

$$|g|\mid 27.$$

The positive divisors of 27 are 1, 3, 9, and 27. Since $g$ is not the identity, its order is not 1. Thus $|g|$ is one of 3, 9, or 27.

If $|g|=3$, then $g$ itself has order 3. If $|g|=9$, then $g^3$ has order 3. If $|g|=27$, then $g^9$ has order 3. In every case, there is an element $a$ of order 3, and its generated subgroup

$$\langle a\rangle=\{e,a,a^2\}$$

has exactly three elements. That proves the claim.

Why the powers have order 3

For an element $g$ of finite order $n$, the order of $g^k$ is

$$\frac{n}{\gcd(n,k)}.$$

In the two cases where we take a power, this gives

$$|g^3|=\frac{9}{\gcd(9,3)}=3$$

when $|g|=9$, and

$$|g^9|=\frac{27}{\gcd(27,9)}=3$$

when $|g|=27$. These calculations explain why those powers generate subgroups of order 3.

A concrete example

Consider the cyclic group $G=\mathbb{Z}/27\mathbb{Z}$ under addition. The element $1$ has order 27. Taking nine times that element gives $9$, whose additive order is 3: the elements generated by $9$ are

$$\langle 9\rangle=\{0,9,18\}.$$

This is a subgroup with three elements. The same conclusion holds for every group of order 27, even if the group is not cyclic or not abelian; the proof depends only on Lagrange’s theorem and the possible element orders.

The connection to Cauchy’s theorem

Cauchy’s theorem says that if a prime $p$ divides the order of a finite group, then the group contains an element of order $p$. Here $3$ divides $27$, so Cauchy’s theorem directly guarantees an element of order 3. The argument above is a particularly short proof for groups whose order is a power of 3: any nonidentity element has order a nontrivial power of 3, and a suitable power of it has order 3.

More generally, if a finite group has order $p^n$ for a prime $p$, then it has a subgroup of order $p$. Choose a nonidentity element of order $p^k$; raising it to the power $p^{k-1}$ produces an element of order $p$.

Common mistakes

• Assuming the group must be cyclic. A group of order 27 need not be cyclic. The proof works for every group of that order.
• Thinking the chosen element must already have order 3. It might have order 9 or 27; taking a power produces an element of order 3.
• Confusing an element with its generated subgroup. An element of order 3 generates a subgroup with three elements, including the identity. The subgroup is not just the set containing that element.
• Using Lagrange’s theorem in the wrong direction. Lagrange says subgroup orders divide the group order; it does not say that every divisor must occur as a subgroup order. The power argument (or Cauchy’s theorem) supplies the existence here.