$a$ and $b$ are elements of the group $G$, then if $a \in \langle b \rangle$, then $\langle a\rangle \subseteq \langle b\rangle$

Question

Assume that $a$ and $b$ are elements of the group $G$, then if $a \in \langle b\rangle $, then $\langle a\rangle \subseteq \langle b\rangle$

How can I prove the foregoing identity? It is from the book modern algebra exercise $27$. It doesn't have solution in the book. Can you help me?

The only thing came to my mind:

$a$ is element of the set generated by $b$. If we say that the order of $\langle b\rangle$ is $k$, then $a^{xy}=b^k=e$. This knowledge doesn't help me

Any link,answer or additional material appreciated !

Answer 1

We have a group $(G,\cdot)$ Let's have a look at the definitions $$\langle b \rangle := \{g \in G : g = b^k, k\in \Bbb Z\}\\ \langle a \rangle := \{g \in G : g = a^h, h\in \Bbb Z\}$$

So we have that if $g \in \langle a \rangle $ $$g = a^h$$ For the hypothesis $a\in \langle b\rangle$ so we have $$g = a^h = (b^k)^h = b^{kh}\in \langle b \rangle$$ This argument holds for every $g \in \langle a \rangle$ and so $\langle a \rangle \subseteq \langle b \rangle$.

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P.S. I'm using multiplicative group notation could you prove the same holds in additive notation?

Answer 2

Note that $\langle x\rangle$ is all powers of $x$. If $a$ is a power of $b$, then all powers of $a$ are powers of $b$.

Answer 3

You really want to take this at a much more abstract level.

If $S$ is a subset of a group $G$, then $\langle S\rangle$ has the following universal property:

1. $\langle S\rangle$ is a subgroup; and
2. $S\subseteq \langle S\rangle$; and
3. If $H$ is any subgroup of $G$ such that $S\subseteq H$, then $\langle S\rangle\subseteq H$.

In fact, $\langle S \rangle$ can be defined as the unique subgroup of $G$ having these three properties (this is the "top-down definition"; you'll also want the "bottoms-up definition" of $\langle S\rangle$, which describes what the elements look like: products of powers of elements of $S$; see here for a general discussion of these types of constructions).

Given that, if $a\in \langle b\rangle$, then since $\langle b\rangle$ is a subgroup, and $a\in \langle b\rangle$, it follows by property 3 above that $\langle a\rangle\subseteq \langle b\rangle$. Presto, done.