Exercise 2153Algebra › GroupsM1Added 3 Nov 2026

Sylow's first theorem

#Sylow#p-subgroup#existence

Statement

State Sylow's first theorem and apply it to a group of order $12$ .

Hint

For $|G|=p^a m$ with $\gcd(p,m)=1$ , $G$ has a subgroup of order $p^a$ .

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Solution

Statement. Let $G$ be a finite group of order $p^a m$ with $p\nmid m$ . Then $G$ has a subgroup of order $p^a$ (a Sylow $p$ -subgroup).

Application to $|G|=12$ . $12=2^2\cdot3$ . Sylow's first theorem gives:
- A subgroup of order $4$ (Sylow $2$ -subgroup).
- A subgroup of order $3$ (Sylow $3$ -subgroup).

Conclusion. Every group of order $12$ has subgroups of order $4$ and $3$ .

Sylow's first theorem

Statement

Solution

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