Question about Abstract Algebra Notation for Abelian Group I've never used this forum before but I had a question about the formatting of one my homework problems.
Prove: If $H$ is an Abelian subgroup of a group $G$, then $\langle H, Z(G)\rangle$ is an Abelian group.
I suppose my confusion is with the use of generator notation containing two arguments.
I understand that the problem is essentially asking me to show if we have an Abelian subgroup $H$ and the center of $G,$ some combination of their elements must be Abelian.
In some kind of psuedo-proof logic:

*

*We know for all $x,y$ in $H$, $xy=yx$

*We know for all $x$ in $Z(G)$ and all $y$ in $G$, $xy=yx$
Is $\langle H, Z(G)\rangle$ a generator for a group with elements that are pairs $(x,y)$?
Thank you in advance if you can help me in any way. I looked through D&F and online but wasn't able to find much or wasn't equipped mentally to know what to look for.
 A: This is a slightly more rigorous look at how you’d prove this.
Given any group $G$ and $S\subseteq G,$ we can define $\langle S\rangle$ as the smallest subgroup of $H$ which contains $S.$ “Smallest” is a bit vague, but we can write it as:

Let$$G_S=\{H\mid H\text{ a subgroup of }G\text{ and }S\subseteq H\}.$$ This set is non-empty since $G\in G_S.$Then define $$\langle S\rangle =\bigcap_{H\in G_S} H$$
Any intersection of subgroups of $G$ is a subgroup of $G.$

This definition has a drawback. It is hard to prove things about $\langle S\rangle.$
We have an equivalent inductive definition:

Given $S,$ define $$S_0=\{1\}\cup S\cup \{s^{-1}\mid s\in S\}$$ and then recursively: $$S_{n+1}=\{s_1s_2\mid s_1,s_2\in S_n\}$$ Since $1\in S_0,$ we get $S_n\subseteq S_{n+1}.$ Then define: $$\langle S\rangle =\bigcup_{n=0}^{\infty} S_n$$

You can prove this subset of $G$ is a group, because you can prove inductively that if $s\in S_n$ then $s^{-1}\in S_n,$ and if $s_1\in S_m, s_2\in S_n$ then $s_1s_2\in S_{\max(m,n)+1}.$
This second definition is more intuitive if you call $\langle S\rangle$ the subgroup “generated by $S.$” We see it “generated” step by step.
You can also prove this is equivalent to the first definition by showing that if $S\subseteq H$ for some subgroup $H$ then, inductively, each $S_n\subseteq H,$ so $\langle S\rangle \subseteq H.$
This second definition is how you’ll prove $\langle H,Z(G)\rangle=\langle H\cup Z(G)\rangle$ is abelian.
More generally:

If every pair of elements $s_1,s_2\in S$ commute, then $\langle S\rangle$ is abelian.

The proof of this is to prove it inductively for any $s_1,s_2\in S_n.$ Note the base case $n=0$ requres a tiny bit of work because you have deal with the inverses. Not hard, but some work.
The induction step is relatively easy.
You’ve already proved every pair of elements of $S=H\cup Z(G)$ commutes, so you are done.
A: Thank you all so much.
Here was my first stab at this proof.
Let G be a group.
Let H be an Abelian subgroup of G.
By the definition of the center of a group, Z(G) = {x∈G|∀y∈G (xy=yx)}
The generator <H, Z(G)> = <H ∪ Z(G)> = H ∪ Z ∪ {hz | h∈H, z∈Z(G)}
Let x,y ∈ H. Because H ≤ G, x,y ∈ G.
Any element z∈Z(G) commutes with all elements of G => zx = xz and zy=yz
Because all elements in the set generated by <H, Z(G)> commute, then <H, Z(G)> is Abelian.
