MHB Proving That $G$ is Abelian When $G/Z$ is Cyclic

[mathmari]

Hey!

Let $Z\subseteq Z(G)$ such that $G/Z$ is cyclic.

I want to show that $G$ is abelian. We have the following:

$$Z(G)=\{g\in G\mid ga=ag \ \forall a \in G\} \\ G/Z=\{gz\mid g\in G\}, z\in Z$$

Since $G/Z$ is cyclic we have that $(gz)^n=1$.

To show that $G$ is abelian, we want to show that $g_1g_2=g_2g_1, \ \forall g_1, g_2\in G$.

Does it stand that since $g_1\in G$ and $g_2\in G$, then $g_1g_2\in G$ ? (Wondering)

From the fact that $G/Z$ is cyclic we have that $((g_1g_2)z)^n=1$ and $(g_iz)^n=1, i\in \{1,2\}$.

Since $z\in Z\subseteq Z(G)$, we have that $za=az, \forall a\in G$, so $(g_1g_2)z=z(g_1g_2)$.

Therefore, from $((g_1g_2)z)^n=1$ we get that $(g_1g_2)^nz^n=1$, right? (Wondering)

And from $(g_1z)^n=1$ we get that $g_1^nz^n=1$.

So, we have the following $$(g_1g_2)^nz^n=1=g_1^nz^n \\ \Rightarrow (g_1g_2)^n=g_1^n \\ \Rightarrow g_1g_2(g_1g_2)^{n-1}=g_1g_1^{n-1} \\ \Rightarrow g_2(g_1g_2)^{n-1}=g_1^{n-1} \\ \Rightarrow g_2(g_1g_2)^{n-1}g_1=g_1^{n-1}g_1 \\ \Rightarrow (g_2g_1)^n=g_1^n$$ So, we have that $$ (g_1g_2)^n=(g_2g_1)^n$$

Is this correct so far? (Wondering)

 

[Deveno]

You are taking the long way around, and may not get there, since to show:

$(g_1g_2)^k = (g_2g_1)^k$ for $1 \leq k < n$ you'll need to use commutativity.

This is what I recommend:

$G/Z$ is a PARTITION of $G$, that is, every element of $G$ can be written in the form $gz$, where $z \in Z$.

So pick $g \in G$ such that $gZ$ is a generator of $G/Z$.

Then, for any $x,y \in G$, we can write:

$x = g^kz_1$, and $y = g^mz_2$ (any $x,y$ fall into one of the cosets $(gZ)^t = g^tZ$ for some $t$).

Can you continue (use the fact that $z_1,z_2 \in Z$ to move them around as needed)?

 

[mathmari]

This is what I recommend:

$G/Z$ is a PARTITION of $G$, that is, every element of $G$ can be written in the form $gz$, where $z \in Z$.

Why is $G/Z$ a partition of $G$ ? And why can every element of $G$ be written in that way? I got stuck right now... (Wondering)

So pick $g \in G$ such that $gZ$ is a generator of $G/Z$.

$G/Z$ has a generator because it is cyclic, right? (Wondering)

Then, for any $x,y \in G$, we can write:

$x = g^kz_1$, and $y = g^mz_2$ (any $x,y$ fall into one of the cosets $(gZ)^t = g^tZ$ for some $t$).

Why does this stand? I haven't understood that. Could you explain it to me? (Wondering)

 

[Deveno]

The relation:

$g_1 \sim g_2 \iff g_1H = g_2H$ is an equivalence relation, called "congruence modulo $H$" (actually, it should be called (left) congruence-but for normal subgroups left and right congruence modulo $H$ are the same).

A *partition* of a set $S$ is a collection of subsets $\{B_i\}$ such that:

$\bigcup\limits_i B_i = S$ and $B_{i_1} \cap B_{i_2} = \emptyset$ when $i_1 \neq i_2$ ($S$ is a union of disjoint sets).

A basic result of algebra is that every equivalence relation on a set $S$ induces a partition-the equivalence classes.

We can see that the union of all equivalence classes is the entire set, since for every $x \in S$, we have $x \in [x]$, since an equivalence relation is reflexive.

Suppose $[x] \neq [y]$ for two $x,y \in S$. I will show $[x]\cap[y] = \emptyset$. For suppose not, that we have $z\in S$ with $z \in [x] \cap [y]$. Since $z \in [x]$, we have (by definition of an equivalence class:

$[x] = \{s \in S: x\sim s\}$)

$x \sim z$.

Since $z \in [y]$, we also have $y \sim z$.

Since equivalence relations are symmetric, from $y \sim z$, we have $z \sim y$. Since equivalence relations are transitive, from:

$x \sim z$ and $z \sim y$, we have $x \sim y$. Thus $y \in x$.

Now if $a \in [y]$, so that $y \sim a$, we have $x \sim y$ and $y \sim a$ so $x \sim a$, and thus $a \in [x]$, which shows that if $[x] \cap [y] \neq \emptyset$, then $[y] \subseteq [x]$.

On the other hand, if $b \in [x]$, so that $x \sim b$, then $b \sim x$, and $x \sim y$, so that $b \sim y$, so $y \sim b$ and thus $[x] \subseteq [y]$.

That is, if $z \in [x] \cap [y]$, then $[x] = [y]$. By supposition, we held $[x] \neq [y]$, contradiction. That is, if $[x] \neq [y]$, no such $z \in [x] \cap [y]$ can exist, so it must be that the intersection is empty.

It is easy to show that "congruence modulo $H$" is an equivalence relation for a group $G$ with a subgroup $H$:

$g \sim g$ for all $g \in G$, since $gH = gH$.

If $g_1 \sim g_2$, then $g_1H = g_2H$, thus $g_2H = g_1H$, so $g_2 \sim g_1$ (it is symmetric).

Finally, if $g_1H = g_2H$ and $g_2H = g_3H$, then (by transitivity of equality), $g_1H = g_3H$.

This is what I mean by a quotient group $G/Z$ PARTITIONS $G$. Every element lies in exactly ONE (and only one) coset of $Z$. I would play around with some groups of small order to see how this partitioning works out in practice.

You are correct, $G/Z$ has a generator because it is cyclic.

The cosets of $G/Z$ for a generator $gZ$, will be, explicitly:

$\{Z,gZ,g^2Z,\dots,g^{n-1}Z\}$ (I'm using "your" $n$).

Every element of $G$, like $x$, for example, will lie in $g^tZ$ for some (unique) $0 \leq t < n-1$.

An element of $g^kZ$ will be of the form $g^kz$, for some $z \in Z$. Note I don't say "which" $z$, because it turns out that it won't matter, the only property of $Z$ you will need is that all its elements commute with any other elements of $G$ (since $Z \subseteq Z(G)$).

 

[mathmari]

I think I got it but I have to think about it again... (Thinking)

$x = g^kz_1$, and $y = g^mz_2$ (any $x,y$ fall into one of the cosets $(gZ)^t = g^tZ$ for some $t$).

Why does it stand that $(gZ)^t = g^tZ$ ? (Wondering)

Can you continue (use the fact that $z_1,z_2 \in Z$ to move them around as needed)?

We have the following:

$$xy=g^kz_1g^mz_2$$

Since $z_1,z_2\in Z\subseteq Z(G)$ we have that $g^iz_1=z_1g^i$ and $g^iz_2=z_2g^i$ for all $i$.

Then $$xy=g^kz_1g^mz_2=g^kz_1z_2g^m=g^kz_2z_1g^m=z_2g^kg^mz_1=z_mg^{k+m}z_1=z_2g^{m+k}z_1=z_2g^mg^kz_1=g^mz_2g^kz_1=yx$$

Theregore, $G$ is abelian. Is this correct? (Wondering)

 

[Deveno]

I think I got it but I have to think about it again... (Thinking)


Why does it stand that $(gZ)^t = g^tZ$ ? (Wondering)

We multiply cosets like this:

$(aH)(bH) = (ab)H$.

Thus:

$(gZ)^2 = (gZ)(gZ) = g^2Z$, and so on...

We have the following:

$$xy=g^kz_1g^mz_2$$

Since $z_1,z_2\in Z\subseteq Z(G)$ we have that $g^iz_1=z_1g^i$ and $g^iz_2=z_2g^i$ for all $i$.

Then $$xy=g^kz_1g^mz_2=g^kz_1z_2g^m=g^kz_2z_1g^m=z_2g^kg^mz_1=z_mg^{k+m}z_1=z_2g^{m+k}z_1=z_2g^mg^kz_1=g^mz_2g^kz_1=yx$$

Theregore, $G$ is abelian. Is this correct? (Wondering)

Yes, that is correct.

 

[mathmari]

Ok... Thanks a lot! (Yes)