Let $G$ be a group with order $14$

Question

Let $G$ be a group with order $14$. Proof that:
a) $G$ contains normal proper subgroup
b) $G$ is isomorphic with $\mathbb{Z}_{14}$ or $G$ is isomorphic with $D_7$.

Solution:

$H \le G$ is normal $ \Leftrightarrow \forall_{g\in G} g H g^{-1} = \{ ghg^{-1} : h \in H \} = H$

From Lagrange theorem: $|G| < \infty, H \le G \implies |G| = |H| \cdot [G:H]$

From Cauchy theorem: $|G| < \infty$, p is prime, $p\mid | G| \implies $ in $G$ exists an element with order $p$, so also subgroup with order $p$

Prime divisions of $14:2,7$ so in $G$ exists $|H_1| = 2, |H_2| = 7$. In each of these subgroups exists neutral element and some different elements so at total we have $1+1+6=8$. So we don't have still considered $6$ element what give us $2$ possibilities for $2$ subgroups with order $7$, one with order $2$ or four with order $2$, one with order $7$.

But how can I proceed with that?

Answer 1

Hint: The number of $p$-Sylows divides $\frac{|G|}{p^v}$, if $v$ is the highest power of $p$ which divides $|G|$.

Answer 2

Since $H_2$ has index 2, it is a normal subgroup.

For the second part, take $s \in H_1$ and $r \in H_2$. If $s$ and $r$ commute, then $$ G \cong \langle s \rangle \times \langle r \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_7 \cong \mathbb{Z}_{14}. $$ If $s$ and $r$ don't commute, you can use this answer anomaly provided.

Edit: For a more elementary solution of the second part, recall that $D_7$ is the unique group generated by some $\sigma$ and $\rho$ such that $\rho^7 = \sigma^2 = 1$ and $\sigma \rho \sigma = \rho^{-1}$. So it suffices to check that $r, s \in G$ statisfy the same relations if they don't commute.

Since $H_2$ is normal, we have $srs = srs^{-1} \in H_2 = \langle r \rangle$, thus $srs = r^n$ for some $n \in \mathbb{Z}$. Rewriting, we get $r = sr^ns = (srs)^n = r^{n^2}$. Since $r$ has order $7$, we have $n^2 \equiv 1$ mod $7$, which has as only solutions $n \equiv \pm 1$. But if $n \equiv 1$, then $srs = r$ so $sr = rs$, contrary to the assumption that $r$ and $s$ don't commute.

So we get $srs = r^{-1}$, as required. If you're not convinced this is sufficient, you can just make an explicit isomorphism $\phi \colon G \to D_7 \colon s^k r^l \mapsto \sigma^k \rho^l$. We can express every $g \in G$ as $s^kr^l$ for some $k, l \in \mathbb{Z}$, since $|H_1 H_2| = \frac{|H_1 | |H_2|}{|H_1 \cap H_2|} = 14 = |G|$, so this is well defined. Also, $D_7$ is generated by $\sigma$ and $\rho$, thus $\phi$ is surjective, and since $|G| = |D_7| = 14$ it must be injective as well.