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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 !

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    You cannot generally say the order of $b$ is $k$, because $k$ may be infinite! – Al.G. Nov 25 '23 at 19:03
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    Forget about order. Write down the definitions: What are you given and what must you prove? – Ted Shifrin Nov 25 '23 at 19:04
  • @TedShifrin I want to prove the title –  Nov 25 '23 at 19:06
  • @TedShifrin which definitions ? this is just an exercise in the book –  Nov 25 '23 at 19:06
  • @randoman Ted Shifrin means: what is the definition of $\langle x\rangle$? The reason for this question is that (a) there are multiple (equivalent) ways to define this notation, so it's helpful to know which you're working with, and (b) like many exercises at the beginning of group theory, this is an example of an exercise which is very easy once you have written out the definitions. – Alex Kruckman Nov 25 '23 at 19:14
  • $a=b^k$ for some $k$. Let $z\in\mathbb Z$, then $a^z=(b^k)^z=b^{kz}$. – Deif Nov 25 '23 at 19:33
  • Duplicate of many prior questions - should not have been answered (or asked). – Bill Dubuque Nov 25 '23 at 21:22
  • @BillDubuque look at your own bussiness, not prevent my education –  Nov 26 '23 at 08:53

3 Answers3

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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.

Shaun
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Arturo Magidin
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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 $\langle a \rangle \subseteq \langle b \rangle$.

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

Turquoise Tilt
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  • why did we say that $a^h=b^k=g$,, they do not have to generate all elements in G –  Nov 25 '23 at 19:18
  • if $g=a^h=b^k$, then why $a^h = b^{kh}$. This example burned me –  Nov 25 '23 at 19:31
  • You can prove the laws of exponents in any group using mathematical induction, more-or-less as you would prove them in the group $\mathbb R- {0}$ (except not the law $(ab)^n = a^n b^n$ which requires the group to be abelian). – Lee Mosher Nov 25 '23 at 19:42
  • @GeoffreyTrang ups... edited. – Turquoise Tilt Nov 25 '23 at 20:40
  • @randoman $g$ is used to indicate a general element in the group $G$, in the definitions $g$ is not the same element, it's just a notation. Do you need further elaboration? – Turquoise Tilt Nov 25 '23 at 20:42
  • Do you have any idea how many books are called "Modern Algebra"? If you are trying to identify a book, you need to provide both title and author. – Arturo Magidin Nov 25 '23 at 21:06
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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 $b$.

Shaun
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