# Suppose $a \in \langle b\rangle$. Then $\langle a\rangle = \langle b\rangle$ if and only if $a$ and $b$ have the same order.

Suppose $a \in \langle b\rangle$. Then $\langle a\rangle = \langle b\rangle$ if and only if $a$ and $b$ have the same order.

Here is what I have so far.

Suppose that $a$ has order $n$. If $a \in \langle b\rangle$ then $a=b^k$ for some $k \in \mathbb{Z}$.

$\langle a\rangle = \langle b^k\rangle = \{e, b^k, b^{2k}, \dots, b^{k(n-1)}\}$

I don't really know where to take it from here. Any help would be appreciated!

I guess I'm very confused about when two cyclic subgroups are equal and what does that mean exactly. Does that mean they're generated by the same element?

• The claim is true only in the finite case. Oct 22, 2017 at 21:03
• To add onto Hagen's statement, if $\langle a\rangle = \langle b\rangle$, then clearly $a$ and $b$ have the same order (infinite or not). On the other hand, if $a$ and $b$ have infinite order, this is not so since, for example, $4\in \langle 2\rangle$, but clearly $\langle 4\rangle\neq \langle 2\rangle$. If $a$ and $b$ have finite order, then what you described is the correct ideal. All you need to do is show those $b^k$ you described are distinct.
– user494247
Oct 22, 2017 at 21:13

Suppose the group is $S_3$ (permutations on $\{1,2,3\}$); if $a=(123)$ and $b=(132)$, then they both generate the same subgroup: $$\langle (123)\rangle=\langle(132)\rangle=\{\mathit{id},(123),(132)\}$$ (where $\mathit{id}$ denotes the identity permutation). Thus, different elements can generate the same (cyclic) subgroup.

Let's stick with elements of finite order (for instance, the group where they live is finite).

You should know the following statement; if not, prove it.

If $a$ is an element of finite order $n$ in a group $G$, then the subgroup $\langle a\rangle$ has precisely $n$ elements.

Thus, if $\langle a\rangle=\langle b\rangle$, the elements $a$ and $b$ have the same order.

Conversely, if $a$ has the same order as $b$, then $\langle a\rangle$ has the same number of elements as $\langle b\rangle$. But from $a\in\langle b\rangle$ it follows that $\langle a\rangle\subseteq\langle b\rangle$. Therefore…

If the elements have infinite order, the statement doesn't hold in general, so the assumption on finite order is necessary.

• Not OP, but I have a question. Was the problem statement supposed to say instead "Suppose ... Then $\langle a \rangle \cong \langle b \rangle$"? Becuse if these are subgroups of some larger group and $a \neq b$, wouldn't $\exists a \in \langle a \rangle$ such that $a \not \in \langle b\rangle$, mainly $a$ itself? So the sets can't be literally equal then if they hold different elements? Oct 23, 2017 at 0:02
• @AndrewTawfeek In the infinite order case, it can happen, but I am more inclined to think that the context for the OP is “finite groups”. Oct 23, 2017 at 6:17
• Ahh okay, thanks for the reply :) Oct 23, 2017 at 6:19

Subgroups are equal if they are equal as sets. So you need to show that

$$\{ a^m : m \in \mathbb{Z} \} = \{ b^n : n \in \mathbb{Z} \}.$$

So then, yes, if they are generated by the same element then they are the same subgroup, but there are easier ways to conclude they are equal, like the equivalence you're trying to prove here.

# Some Suggestions

It sounds like you're working with finite groups. Here are some tips to get you started:

• Assume that $a \in \langle b \rangle$ and that $n = \operatorname{order}(b) \in \mathbb{N}$. This can be done when proving both implications. You are correct that you can also assume $a = b^k$ for some $k \in \mathbb{N}$.

• [For $\implies$] What do you know about the size of the set $\langle a \rangle$ given that $a \in \langle b \rangle$? How does the order of $a$ relate to the size of $\langle a \rangle$? Use this to make a conclusion in the case where $\langle a \rangle = \langle b \rangle$.

• [For $\impliedby$] If $\operatorname{order}(a) = n$, write $\langle a \rangle$ like you did in your question: $$\langle a \rangle = \{1, b^k, b^{2k}, \ldots, b^{(n-1)k}\}.$$ Given that $b^{rk} \neq b^{sk}$ when $0 \leq r,s < n$, you can conclude something about how $k$ and $n$ are related. This can be used to prove the reverse implication.