Proving two groups of the same order are equal?

Question

Two finite (cyclic) groups $\langle a\rangle$ and $\langle x\rangle$ have the same order. My professor said that instead of showing that $\langle a\rangle \subseteq \langle x\rangle$ and $\langle x\rangle \subseteq \langle a\rangle$ to prove that $\langle a\rangle = \langle x\rangle$, I just need to show either $\langle a\rangle \subseteq \langle x\rangle$ or $\langle x\rangle \subseteq \langle a\rangle$ because the groups are the same order? Why is this?

Answer 1

I think your professor just means that since the statement is symmetric $a$ and $x$, if you manage to show $\langle a \rangle \subset \langle x \rangle$, then switching the roles of $a$ and $x$, you obtain the other inclusion, and thus the equality.

Answer 2

It is enough to prove that $a\in\left\langle x\right\rangle $ (or $x\in\left\langle a\right\rangle $) . This implies that $\left\langle a\right\rangle $ is a subgroup of $\left\langle x\right\rangle $. It has the same finite cardinality as $\left\langle x\right\rangle $ because $a$ and $x$ have the same finite order. So $\left\langle a\right\rangle =\left\langle x\right\rangle $

Answer 3

There are two things going on here. The first is that, if $f: S\to T$ is an injective function between finite sets, and $|S| = |T|$, then $f$ is a bijection. This should be intuitively clear: since $f$ is injective, the image of $f$ must have the same size as $S$, so $f$ cannot miss any elements. It is good practice to try writing a rigorous proof, though.

The second thing is that a group homomorphism which is a bijection must also have an inverse that is itself a group homomorphism. This is subtle, since the same theorem is not true for all mathematical structures. It fails for topological spaces, for example.

So let's be explicit about it:

Theorem: If $f: G\to H$ is a bijective group homomorphism, then it is an isomorphism.

Proof: Define $f': H\to G$ to be the inverse of $f$, as a map of sets. In other words, $f'(h)$ is defined to be the unique $g\in G$ with $f(g) = h$.

It is clear that $f\circ f' = \operatorname{id}_H$, and $f'\circ f = \operatorname{id}_G$. So the only thing that we need to check is that $f'$ is a homomorphism of groups.

So, take some $h_1,h_2\in H$, and suppose that $f'(h_1) = g_1$ and $f'(h_2) = g_2$. By definition, $f(g_1) = h_1$ and $f(g_2) = h_2$. So, since $f$ is a group homomorphism, we have $f(g_1 g_2) = f(g_1)\cdot f(g_2) = h_1 h_2$, and, by definition, $f' (h_1 h_2) = g_1 g_2 = f'(h_1)\cdot f'(h_2)$.

We conclude that $f'$ is an inverse for $f$, so $f$ is an isomorphism.

Answer 4

Equivalently, if $|G|=n$ is cyclic then $$G\cong\mathbb Z_n$$ Note that $\cong$ is an equivalence relation among groups.