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I am reading a chapter on Commutator in Group Theory and came across chevron symbols "$\langle$" and "$\rangle$" like these:

Question #1: Let $E$ and $F$ be non-empty subsets of $G$, we set $$[E, F] := \langle [e,f] \mid e \in F, f \in F \rangle.$$

Correct me if I am wrong here: The chevron symbols here do not have special meaning except that they are differentiating from "$\{$" and "$\}.$" The text does not give any explanation.

And then on the same page I saw this problem:

Question #2: Let $H$ be a subgroup of $G.$ Show that $[H, g] = [H, \langle g \rangle]$ for each element $g$ in $G.$

Here the chevron signs have special meaning, they refer to $g$ as generator. If I am correct, how do you go about solving this problem? My understanding that I have to start from the LHS, without assuming that $g$ is a generator. Please help and thanks for your time.

POST SCRIPT - 1: ~~~~~~~~~~~~~~~~~~~~~~
I am following hints from "someone you know" in solving Question #2 like these:

$$\begin{align*}[H,g] &= \{hgh^{-1}g^{-1} \mid h \in H \}\\ &= \{hg (g^{n-1})(g^{n-1})^{-1}h^{-1}g^{-1} \mid h \in H \} \\ &= \{h(g g^{n-1})(g^{n-1})^{-1}h^{-1}g^{-1} \mid h \in H \} \\ &= \{hg^n(g^{n-1})^{-1}h^{-1}g^{-1} \mid h \in H \} \\ &= ... \\ &= ... \\ &= [H, \langle g \rangle] \end{align*}\\$$

How do you go from the 4th. line to the next to move the $(g^n)^{-1}$ to the right side of $h^{-1}$ without resorting to the group being abelian? Thank you again.

POST SCRIPT - 2: ~~~~~~~~~~~~~~~~~~~~~~
Since I did not receive any feedback after the Post Script - 1, and since on the subsequent pages of the same class note I saw a very similar question like this:

Let $G$ be a group and let $g \in G,$ and let $N$ be commutative normal subgroup of $G.$ Show that $[N, \langle g \rangle] = \{[n, g] \} \mid n \in N \}.$

Do you agree with me that there is actually a typo in the Question #2, in that $G$ and therefore $H$ should be declared as commutative in the first place? Thanks for your time and feedback.

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  • $\begingroup$ By "they refer to $g$ as [a] generator" you mean "$\langle g\rangle$ refers to the cyclic subgroup generated by $g$." $\endgroup$
    – anon
    Dec 23, 2014 at 22:34
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    $\begingroup$ Another name for $\langle \rangle$ is angle brackets. $\endgroup$ Dec 24, 2014 at 14:01

2 Answers 2

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In general, if $S$ is any subset of a group $G$, the notation $\langle S\rangle$ refers to the subgroup generated by the elements of $S$, so $$[E, F] := \langle [e,f] \mid e \in F, f \in F \rangle$$ is the subgroup of $G$ generated by commutators coming from elements of $E$ and $F$. Thus, the notation $\langle g\rangle$ is a special case, in that it refers to $\langle \{g\}\rangle$.

The problem you refer to is asking you to show that $\langle S_1\rangle=\langle S_2\rangle$ where $$\begin{align*} S_1&=\{hgh^{-1}g^{-1}:h\in H\}\\ S_2&=\{hkh^{-1}k^{-1}:h\in H,k\in\langle g\rangle\}=\{hg^nh^{-1}g^{-n}:h\in H, n\in\mathbb{Z}\} \end{align*}$$

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  • $\begingroup$ So, the $\langle g \rangle$ refers to generator? Thanks again. $\endgroup$
    – A.Magnus
    Dec 23, 2014 at 22:25
  • $\begingroup$ It means "generated by" rather. $\endgroup$ Dec 24, 2014 at 14:23
  • $\begingroup$ @someone you know : To "Someone You Know": Could you please help me with the Post Script - 2 above? To all other heavyweights beside "Someone You Know," you are more than welcome to pitch in. Thanks again. $\endgroup$
    – A.Magnus
    Dec 26, 2014 at 4:25
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(1). $[E, F] := \langle [e,f] \mid e \in F, f \in F \rangle$ means the subgroup of $G$ generated by the elements of the form $[e,f], e \in E, f \in F.$

(2). $\langle g\rangle$ means the subgroup of $G$ generated by the element $g.$

You can look at here or at any algebra text book.

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