Elements Order 3 C984B5

1. **Problem Statement:** We want to prove that a noncyclic group $G$ of order $21$ must have $14$ elements of order $3$. 2. **Recall the structure and relevant theorems:** By the Sylow theorems, the number of Sylow $p$-subgroups, denoted $n_p$, satisfies: - $n_p \equiv 1 \pmod p$ - $n_p$ divides the order of the group divided by $p^k$, where $p^k$ is the highest power of $p$ dividing the group order. 3. **Apply Sylow theorems to $G$ of order $21 = 3 \times 7$:** - Number of Sylow $7$-subgroups, $n_7$, divides $3$ and $n_7 \equiv 1 \pmod 7$. The divisors of $3$ are $1$ and $3$. Only $1$ satisfies $n_7 \equiv 1 \pmod 7$. Thus, $n_7 = 1$. - Number of Sylow $3$-subgroups, $n_3$, divides $7$ and $n_3 \equiv 1 \pmod 3$. The divisors of $7$ are $1$ and $7$. Both satisfy $n_3 \equiv 1 \pmod 3$ since $1 \equiv 1$ and $7 \equiv 1 \pmod 3$. 4. **Noncyclic condition implies $n_3 \neq 1$:** If $n_3 = 1$, then the Sylow $3$-subgroup is unique and normal, and combined with the unique Sylow $7$-subgroup, the group would be cyclic (since it would be isomorphic to the direct product of cyclic groups of orders $3$ and $7$). Since $G$ is noncyclic, $n_3 = 7$. 5. **Count elements of order $3$:** Each Sylow $3$-subgroup has order $3$ and is cyclic, so it has $2$ elements of order $3$ (excluding the identity). Since there are $7$ distinct Sylow $3$-subgroups and they intersect only at the identity element, the total number of elements of order $3$ is: $$7 \times 2 = 14.$$