How Many Automorphisms Does a Cyclic Group of Order 24 Have?

An isomorphism from a group to itself is called an automorphism. For a cyclic group of order 24, count the possible images of one generator: each automorphism is determined by that image, and the image must itself generate the group. This gives a direct count using Euler’s totient function.

The answer: 8 automorphisms

Let $G=\langle g\rangle$ be cyclic of order 24. Every element of $G$ is a power of $g$, and a homomorphism $f:G\to G$ is completely determined by the value of $f(g)$. For the map to be an isomorphism, $f(g)$ must be a generator of $G$.

The powers of $g$ that generate $G$ are precisely $g^k$ for exponents $k$ relatively prime to 24. Therefore the number of automorphisms is

$$|\operatorname{Aut}(G)|=\varphi(24)=24\left(1-\frac12\right)\left(1-\frac13\right)=8.$$

So there are 8 isomorphisms of $G$ onto itself.

Why the image of a generator matters

A homomorphism preserves powers: if $f(g)=a$, then $f(g^n)=a^n$. Since every element of $G$ is a power of $g$, specifying $f(g)$ specifies the whole map. In particular, choosing $f(g)=g^k$ gives the map

$$f_k(g^n)=g^{kn}.$$

This map is an automorphism exactly when $g^k$ has order 24. The order of a power in a cyclic group is

$$\operatorname{ord}(g^k)=\frac{24}{\gcd(24,k)}.$$

Thus $g^k$ has order 24 if and only if $\gcd(24,k)=1$. If the gcd is greater than 1, the image generates only a proper subgroup, so the map is not onto and cannot be an isomorphism.

List the eight maps

Among the exponents $0,1,\ldots,23$, the ones relatively prime to 24 are

$$1,\;5,\;7,\;11,\;13,\;17,\;19,\;23.$$

They give the eight automorphisms by sending $g$ respectively to

$$g,\;g^5,\;g^7,\;g^{11},\;g^{13},\;g^{17},\;g^{19},\;g^{23}.$$

For example, the map $f(g)=g^5$ is an automorphism because $\gcd(5,24)=1$. Its inverse sends $g$ to $g^{29}$, which is the same as $g^5$ because $29\equiv 5\pmod{24}$ and $5\cdot 5=25\equiv1\pmod{24}$. Indeed, applying the map twice sends $g$ to $g^{25}=g$.

A second way to count

The prime factorization is $24=2^3\cdot3$. Euler’s totient function counts integers from 1 to 24 that are coprime to 24, using

$$\varphi(n)=n\prod_{p\mid n}\left(1-\frac1p\right),$$

where the product runs over the distinct prime divisors of $n$. Hence

$$\varphi(24)=24\left(1-\frac12\right)\left(1-\frac13\right)=24\cdot\frac12\cdot\frac23=8.$$

This count agrees with the explicit list of valid exponents.

Common mistakes

• Counting every possible image of $g$. There are 24 possible images, but most do not generate $G$. Only the 8 generators can be images under an automorphism.
• Counting elements of order dividing 24 as generators. Every element of this group has order dividing 24, but a generator must have order exactly 24.
• Confusing automorphisms with group elements. The group has 24 elements; the number of its automorphisms is 8.
• Including exponent 0. The element $g^0=e$ does not generate a group of order 24, so the map sending $g$ to the identity is not an isomorphism.

The same argument works for any finite cyclic group of order $n$: its number of automorphisms is $\varphi(n)$, because the image of a generator must be one of the group’s $\varphi(n)$ generators.