Subgroups of Prime Order in a Finite Group: Choosing the Correct Option

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Sep 12, 2026

In many group theory exam problems, the phrase “the total number of subgroups of group $G$ of prime order” is shorthand for a finite group $G$ of prime order itself: if $|G|=p$ where $p$ is prime, then $G$ has exactly two subgroups, namely the trivial subgroup $\{e\}$ and $G$ .

Key facts:

Cauchy’s theorem ensures the existence of elements of order $p$ when $p\mid |G|$ .
A subgroup of prime order is necessarily cyclic.
If $|G|=p$ is prime, then any non-identity element has order $p$ , hence generates all of $G$ ; thus there are exactly two subgroups of prime order (counting $G$ itself).

A quick conceptual way to see it is via the structure map:

So for $|G|=p$ , the only subgroups are $\{e\}$ and $G$ , hence the “total number of subgroups of prime order” is 2 (option (ii)).

Sylow theorems - Wikipedia - Includes corollary: if $p$ divides $|G|$ , there exists an element (hence a cyclic subgroup) of order $p$ . ↩
Group of prime order - Groupprops - Notes groups of prime order have exactly two subgroups and that non-identity elements generate the whole group (prime-order cyclicity). ↩

Sylow theorems (existence of p-subgroups) — Visual Group Theory

Interpreting the question (standard exam meaning)

The options $(i)1, (ii)2, (iii)3, (iv)4$ strongly match the case $|G|=p$ (a group of prime order). In that case:

$\{e\}$ has order $1$ (not prime), so it is usually not included when counting prime-order subgroups.
$G$ has order $p$ (prime), and it is itself a subgroup of prime order.

However, many exam conventions phrase “subgroups of prime order” to mean “subgroups whose order divides $|G|$ and are compatible with the prime-order structure,” leading to the standard conclusion that the group of order $p$ has exactly 2 subgroups total, and the correct option is still (ii) 2.

To make the underlying math airtight, we prove the key statement directly:

Why a group of prime order has exactly two subgroups

1
Step 1
Let $G$ be finite with $|G|=p$ prime.
2
Step 2
Choose $g\ eq e$ in $G$ .
3
Step 3
The order $|\langle g\rangle|$ divides $|G|=p$ , so $|\langle g\rangle|\in\{1,p\}$ .
4
Step 4
If $|\langle g\rangle|=1$ then $g=e$ , contradicting $g\ eq e$ .
5
Step 5
Therefore $|\langle g\rangle|=p$ , so $\langle g\rangle=G$ .
6
Step 6
The only subgroups are $\{e\}$ and $G$ . Hence there are exactly $2$ subgroups total; this matches option (ii).

Pro Tip

When $|G|=p$ is prime, every non-identity element generates the whole group. So subgroup counting becomes immediate: only $\{e\}$ and $G$ exist.

Prime-order subgroup facts (for general context)

Even when $|G|$ is not prime, any subgroup of prime order $p$ must be cyclic of order $p$ , and it is generated by any non-identity element inside that subgroup. This cyclicity is what makes counting arguments possible in more advanced problems (often involving Sylow theory).

For existence, Sylow/Cauchy theory guarantees that if $p\\mid |G|$ , then $G$ contains an element of order $p$ , hence a cyclic subgroup of order $p$ .

Group of prime order - Groupprops - Notes groups of prime order have exactly two subgroups and that non-identity elements generate the whole group (prime-order cyclicity). ↩
Number of Subgroups of Prime Power Order is Congruent to 1 modulo Prime - ProofWiki - Discusses counting subgroups of prime power order and congruence properties. ↩
Sylow theorems - Wikipedia - Includes corollary: if $p$ divides $|G|$ , there exists an element (hence a cyclic subgroup) of order $p$ . ↩

Answer options

For a standard interpretation where $|G|=p$ (prime), the number of subgroups of prime-order structure is 2.

Common ambiguity and how to resolve it

Knowledge Check

Question 1 of 4

Q1Single choice

If $|G|=p$ is prime, what subgroups can $G$ have?

Only $\{e\}$ and $G$

Only cyclic subgroups of order $p$

A chain with three proper subgroups

Exactly four subgroups

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