Group of Order 30 has Normal Cyclic Subgroup of Order 15

Theorem
Let $G$ be of order $30$.

Then $G$ has a normal subgroup of order $15$ which is cyclic.

Proof
By Group of Order 15 is Cyclic Group, any subgroup of $G$ of order $15$ is cyclic.

It remains to be proved that a subgroup of $G$ of order $15$ exists, and that it is normal.

Let $n_3$ denote the number of Sylow $3$-subgroups of $G$.

From the Fourth Sylow Theorem:
 * $n_3 \equiv 1 \pmod 3$

and from the Fifth Sylow Theorem:
 * $n_3 \divides 30$

where $\divides$ denotes divisibility.

It follows that $n_3 \in \set {1, 10}$.

Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.

From the Fourth Sylow Theorem:
 * $n_5 \equiv 1 \pmod 5$

and from the Fifth Sylow Theorem:
 * $n_5 \divides 30$

It follows that $n_5 \in \set {1, 6}$.

Suppose $n_3 = 1$.

Let $P$ denote the unique Sylow $3$-subgroup of $G$.

From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.

Thus we may form the quotient group $G / P$, which is of order $10$.

From Groups of Order 2p, $G / P$ has a unique Sylow $5$-subgroup, which we will denote $N / P$.

By the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements.

As noted, $N$ is cyclic.

Suppose $n_3 = 10$.

Let the Sylow $3$-subgroups of $G$ be denoted $P_1, P_2, \ldots, P_{10}$.

Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, \ldots, 10}, i \ne j$ is the trivial subgroup of $G$:


 * $P_i \cap P_j = \set e$

Thus $G$ contains:
 * The identity element $e$
 * $20$ elements of order $3$, of which $2$ each are in $P_1, P_2, \ldots, P_{10}$
 * $9$ more elements of $G$ which are in subgroups of $G$ different from the Sylow $3$-subgroups, and so have orders different from $3$.

Suppose $n_5 = 6$.

By a similar argument to the above, these $6$ Sylow $5$-subgroups contribute $24$ elements of order $5$ to $G$.

This is not possible because there are only $9$ elements of $G$ so far unaccounted for.

So if $G$ has $10$ Sylow $3$-subgroups, it can have only $1$ Sylow $5$-subgroup, which we will denote $Q$.

From Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.

Thus we may form the quotient group $G / Q$, which is of order $16$.

From Groups of Order 2p, $G / Q$ has a unique Sylow $3$-subgroup, which we will denote $N / Q$.

Again by the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements.

As noted, $N$ is cyclic.

Both cases have been accounted for:
 * that $G$ has $1$ Sylow $3$-subgroup
 * that $G$ has $10$ Sylow $3$-subgroups.

Both cases result in $G$ having a normal subgroup of order $15$ which is cyclic.