Prove that a group of order 10 is either isomorphic to $\mathbb{Z}_{10}$ or isomorphic to $D_5$

Question

The question is to prove that a group of order 10 is either isomorphic to $\mathbb{Z}_{10}$ or isomorphic to $D_5$. I have been looking for solutions but couldn't find an explanation without Sylow theorems, which i haven't had yet.

Answer 1

Let $G$ be a group of order $10$. Note that by Cauchy's Theorem there are elements of order $2$ and $5$, namely $a^2 = 1, b^5 = 1$. Now as $[G:\langle b \rangle] = 2$ we have that $\langle b \rangle$ is normal in $G$. Therefore we must have:

$$aba^{-1} = b,b^2,b^3 \text{ or }b^4$$

The first case gives us that $G$ is abelian and therefore it's isomorphic to $\mathbb{Z}_{10}$ as $10$ is square-free integer. The second and third case are impossible, as:

$$b = a^2b(a^{-1})^2 = a(aba^{-1})a^{-1} = ab^2a^{-1} = b^4$$ $$b = a^2b(a^{-1})^2 = a(aba^{-1})a^{-1} = ab^3a^{-1} = b^9 = b^4$$

The fourth case is possible. Now show that:

$$(a,b: a^2 = 1, b^5 = 1, ab = ba) \quad \text{ and } \quad (a,b: a^2 = 1, b^5 = 1; ab = b^4a)$$

are two distinct subgroups of order $10$.