Prove that a Group of Order 217 is Cyclic and Find the Number of Generators Problem 458

Let $G$ be a finite group of order $217$.

(a) Prove that $G$ is a cyclic group.

(b) Determine the number of generators of the group $G$. Add to solve later

Sylow’s Theorem

We will use Sylow’s theorem to prove part (a).

For a review of Sylow’s theorem, check out the post “Sylow’s Theorem (summary)“.

Proof.

(a) Prove that $G$ is a cyclic group.

Note the prime factorization $217=7\cdot 31$.
We first determine the number $n_p$ of Sylow $p$-group for $p=7, 31$.
Recall from Sylow’s theorem that
\begin{align*}
&n_p \equiv 1 \pmod{p}\6pt] &n_p \text{ divides } n/p. \end{align*} Thus, n_7 could be 1, 8, 15, 22, 29,\dots and n_7 needs to divide 217/7=31. Hence the only possible value for n_7 is n_7=1. So there is a unique Sylow 7-subgroup P_7 of G. By Sylow’s theorem, the unique Sylow 7-subgroup must be a normal subgroup of G. Similarly, n_{31}=1, 32, \dots and n_{31} must divide 217/31=7, and hence we must have n_{31}=1. Thus G has a unique normal Sylow 31-subgroup P_{31}. Note that these Sylow subgroup have prime order, and hence they are isomorphic to cyclic groups: \[P_7\cong \Zmod{7} \text{ and } P_{31}\cong \Zmod{31}.

It is also straightforward to see that $P_7 \cap P_{31}=\{e\}$, where $e$ is the identity element in $G$.

In summary, we have

1. $P_7, P_{31}$ are normal subgroups of $G$.
2. $P_7 \cap P_{31}=\{e\}$.
3. $|P_7P_{31}|=|G|$.

These yields that $G$ is a direct product of $P_7$ and $P_{31}$, and we obtain
$G=P_7\times P_{31}\cong \Zmod{7} \times \Zmod{31}\cong \Zmod{217}.$ Hence $G$ is a cyclic group.

(b) Determine the number of generators of the group $G$.

Recall that the number of generators of a cyclic group of order $n$ is equal to the number of integers between $1$ and $n$ that are relatively prime to $n$.
Namely, the number of generators is equal to $\phi(n)$, where $\phi$ is the Euler totient function.

By part (a), we know that $G$ is a cyclic group of order $217$.
Thus, the number of generators of $G$ is
\begin{align*}
\phi(217)=\phi(7)\phi(31)=6\cdot 30=180,
\end{align*}
where the first equality follows since $\phi$ is multiplicative. Add to solve later

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1. 06/24/2017

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