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.

  • $\begingroup$ 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 '14 at 22:34
  • 1
    $\begingroup$ Another name for $\langle \rangle$ is angle brackets. $\endgroup$ – Matthew Towers Dec 24 '14 at 14:01

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 $$\begin{align*} S_1&=\{hgh^{-1}g^{-1}:h\in H\}\\ S_2&=\{hkh^{-1}k^{-1}:h\in H,k\in\langle g\rangle\}=\{hg^nh^{-1}g^{-n}:h\in H, n\in\mathbb{Z}\} \end{align*}$$

  • $\begingroup$ So, the $\langle g \rangle$ refers to generator? Thanks again. $\endgroup$ – A.Magnus Dec 23 '14 at 22:25
  • $\begingroup$ It means "generated by" rather. $\endgroup$ – darij grinberg Dec 24 '14 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 '14 at 4:25

(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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.