MHB Show that each such group is solvable

[mathmari]

Hey!

I am looking at the following exercise:

Each group of order $p^2q$, with $p,q$ different primes, is solvable.

To show this can we just say that it holds according to Burnside's Theorem ? (Wondering)

 

[Deveno]

You could...but it is easier to prove it directly, rather than prove Burnside's Theorem, which is a bit "deep" (representation theory is probably a bit beyond what you are studying at the moment).

For simplicity's sake, let $n_p = |\text{Syl}_p(G)|$.

If $|G| = p^2q$ then if $G$ is simple, $n_p = q$.

Also, we have $n_q = p$ or $n_q = p^2$.

Show that $n_q = p^2$ leads to "too many elements of $G$".

So the "hard case" is $n_q = p$.

First, prove this lemma:

Lemma: Let $G$ be a finite group, and $r$ a prime number such that $r$ divides $|G|$. If $G$ is simple, then $n_r > r$.

This should lead you to an easy contradiction.

 

[mathmari]

You could...but it is easier to prove it directly, rather than prove Burnside's Theorem, which is a bit "deep" (representation theory is probably a bit beyond what you are studying at the moment).

For simplicity's sake, let $n_p = |\text{Syl}_p(G)|$.

If $|G| = p^2q$ then if $G$ is simple, $n_p = q$.

Also, we have $n_q = p$ or $n_q = p^2$.

Show that $n_q = p^2$ leads to "too many elements of $G$".

So the "hard case" is $n_q = p$.

First, prove this lemma:

Lemma: Let $G$ be a finite group, and $r$ a prime number such that $r$ divides $|G|$. If $G$ is simple, then $n_r > r$.

This should lead you to an easy contradiction.

In my notes we have proven the lemma:

$|G|=pq^2$ then $G$ is not simple. So, I can use that without proving it again, or not? (Wondering) How can we use the fact that the group is not simple? (Wondering)

 

[Deveno]

If you've read and understood the proof, then sure, go ahead and use it. But..

...since we want to show $G$ is solvable, it would be good to know what normal subgroups we have, not just that we have "one" (or more).

If you follow the steps I have outlined above, you will find that there is a subgroup $N$ such that either:

a) $|G/N| = q$

b) $|G/N| = p^2$.

Either way, the factor group is abelian, so $G$ is solvable (some definitions require the factor groups to be cyclic, or cyclic of prime order, but this poses no real difficulty-why?).

 

[mathmari]

First, prove this lemma:

Lemma: Let $G$ be a finite group, and $r$ a prime number such that $r$ divides $|G|$. If $G$ is simple, then $n_r > r$.

Lemma: Let $G$ be a finite group, and $r$ a prime number such that $r$ divides $|G|$. If $G$ is simple, then $n_r > r$.

Proof:
Since $r$ divides $|G|$, we have that there is a $r$-Sylow in $G$.
$n_r$ divides $|G|$ and is congruent to $1\pmod r$, i.e., $n_r=1+kr$.
That means that either $n_r=1$ or $n_r>r$.
When $n_r=1$ then the Sylow is unique so it is normal in $G$, so $G$ is not simple, a contradiction.
Therefore, $n_r>r$.

Is this correct? (Wondering)

For simplicity's sake, let $n_p = |\text{Syl}_p(G)|$.

If $|G| = p^2q$ then if $G$ is simple, $n_p = q$.

Also, we have $n_q = p$ or $n_q = p^2$.

Show that $n_q = p^2$ leads to "too many elements of $G$".

So the "hard case" is $n_q = p$.

We have that $|G|=p^2q$.
We have that $n_p\mid |G| \Rightarrow n_p=1+np\mid p^2q \Rightarrow n_p\mid q \Rightarrow n_p\in \{1, q\}$ and $n_q\mid |G| \Rightarrow n_q=1+mq\mid p^2q \Rightarrow n_q\mid p^2 \Rightarrow n_q\in \{1, p, p^2\}$.
If $G$ is simple, according to the lemma we have that $n_p>p$ and $n_q>q$.
So, $n_p$ must be equal to $q$, where $q>p$.
Since $n_q>q$ and $q>p$, it cannot be that $n_q=1$ or $n_q=p$. So it must be $n_q=p^2$.
Therefore, we have $p^2$ $q$-Sylow in $G$. That means that we have $p^2(q-1)=p^2q-p^2$ elements of order $q$.
Since $|G|=p^2q$, we can have only one $p$-Sylow ( $|P|=p^2$ ).
So, $n_p=1$, a contradiction.
Therefore, $G$ is not simple.

Is this correct? (Wondering)

If you've read and understood the proof, then sure, go ahead and use it. But..

...since we want to show $G$ is solvable, it would be good to know what normal subgroups we have, not just that we have "one" (or more).

So, we have one $p$-Sylow of order $p^2$, which is normal, right? (Wondering)

If you follow the steps I have outlined above, you will find that there is a subgroup $N$ such that either:

a) |G/N| = q

b) |G/N| = p^2.

Either way, the factor group is abelian, so $G$ is solvable (some definitions require the factor groups to be cyclic, or cyclic of prime order, but this poses no real difficulty-why?).

Having a normal $p$-Sylow of order $p^2$, say $N$, we have that $|G/N|=\frac{|G|}{|N|}=\frac{p^2q}{p^2}=q$, right? (Wondering)

Do we have that each group of order a power of a prime, is abelian? (Wondering)

How do we conclude then that $G$ is solvable? (Wondering)

 

[Deveno]

Lemma: Let $G$ be a finite group, and $r$ a prime number such that $r$ divides $|G|$. If $G$ is simple, then $n_r > r$.

Proof:
Since $r$ divides $|G|$, we have that there is a $r$-Sylow in $G$.
$n_r$ divides $|G|$ and is congruent to $1\pmod r$, i.e., $n_r=1+kr$.
That means that either $n_r=1$ or $n_r>r$.
When $n_r=1$ then the Sylow is unique so it is normal in $G$, so $G$ is not simple, a contradiction.
Therefore, $n_r>r$.

Is this correct? (Wondering)

Yes.

We have that $|G|=p^2q$.
We have that $n_p\mid |G| \Rightarrow n_p=1+np\mid p^2q \Rightarrow n_p\mid q \Rightarrow n_p\in \{1, q\}$ and $n_q\mid |G| \Rightarrow n_q=1+mq\mid p^2q \Rightarrow n_q\mid p^2 \Rightarrow n_q\in \{1, p, p^2\}$.
If $G$ is simple, according to the lemma we have that $n_p>p$ and $n_q>q$.
So, $n_p$ must be equal to $q$, where $q>p$.
Since $n_q>q$ and $q>p$, it cannot be that $n_q=1$ or $n_q=p$. So it must be $n_q=p^2$.
Therefore, we have $p^2$ $q$-Sylow in $G$. That means that we have $p^2(q-1)=p^2q-p^2$ elements of order $q$.
Since $|G|=p^2q$, we can have only one $p$-Sylow ( $|P|=p^2$ ).
So, $n_p=1$, a contradiction.
Therefore, $G$ is not simple.

Is this correct? (Wondering)

Yes.

So, we have one $p$-Sylow of order $p^2$, which is normal, right? (Wondering)

Not necessarily. All we showed above is that $G$ has a normal Sylow subgroup, it might be the $p^2$ one, or the $q$ one. For example, in $A_3$ ($p = 2,q = 3$) it's the Sylow $2$-subgroup that is normal (this group is:

$V = \{e,(1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$), whereas in $D_6$ (the dihedral group of order $12$) it's the Sylow $3$-subgroup:

$\langle r^2 \rangle = \{1,r^2,r^4\}$

that is normal.

Having a normal $p$-Sylow of order $p^2$, say $N$, we have that $|G/N|=\frac{|G|}{|N|}=\frac{p^2q}{p^2}=q$, right? (Wondering)

That is one possibility, you must consider both.

Do we have that each group of order a power of a prime, is abelian? (Wondering)

No, there exist non-abelian groups of order $p^k$ for $k > 2$. But we need not concern ourselves with that. Here, the only two possibilities are:

$|G/N| = q$ (all groups of order $q$ are abelian, in fact, they are cyclic, and simple).

$|G/N| = p^2$ (all groups of order $p^2$ are abelian...why?)

How do we conclude then that $G$ is solvable? (Wondering)

What can you say about the normal series:

$1 \lhd N \lhd G$?

(in either case)?

 

[mathmari]

Not necessarily. All we showed above is that $G$ has a normal Sylow subgroup, it might be the $p^2$ one, or the $q$ one. For example, in $A_3$ ($p = 2,q = 3$) it's the Sylow $2$-subgroup that is normal (this group is:

$V = \{e,(1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$), whereas in $D_6$ (the dihedral group of order $12$) it's the Sylow $3$-subgroup:

$\langle r^2 \rangle = \{1,r^2,r^4\}$

that is normal.

Ah ok... (Thinking)

No, there exist non-abelian groups of order $p^k$ for $k > 2$. But we need not concern ourselves with that. Here, the only two possibilities are:

$|G/N| = q$ (all groups of order $q$ are abelian, in fact, they are cyclic, and simple).

We have that a group of order a prime $q$ is cyclic due to Lagrange's Theorem, right? (Wondering)
We have that the order of each element of the group must divide the the order of the group, $q$.
The divisors of $q$ are $1$ and $q$.
Only the identity has order $1$. So, all the other elements must have order $q$.
Therefore, the group is cyclic, right? (Wondering)

Since $|A|=q$, we have that there are $q$-Sylow in $A$.
We have that $n_q=1+kq$ and $n_q\mid q$. That means that $n_q=1$, or not? (Wondering)

What can you say about the normal series:

$1 \lhd N \lhd G$?

(in either case)?

In both cases, we have that $G/N$ is abelian, as mentioned above and $N/1=N$ is also abelian.

Therefore, $G$ is solvable, right? (Wondering)

 

[Deveno]

Ah ok... (Thinking)

We have that a group of order a prime $q$ is cyclic due to Lagrange's Theorem, right? (Wondering)
We have that the order of each element of the group must divide the the order of the group, $q$.
The divisors of $q$ are $1$ and $q$.
Only the identity has order $1$. So, all the other elements must have order $q$.
Therefore, the group is cyclic, right? (Wondering)

Yep.

Since $|A|=q$, we have that there are $q$-Sylow in $A$.
We have that $n_q=1+kq$ and $n_q\mid q$. That means that $n_q=1$, or not? (Wondering)

I'm not sure what your "$A$" is, here. We're only using the Sylow theory to show one of the Sylow subgroups is normal. This gives us the normal series we need to show $G$ is solvable.

In both cases, we have that $G/N$ is abelian, as mentioned above and $N/1=N$ is also abelian.

Therefore, $G$ is solvable, right? (Wondering)

Indeed, from the definition of solvable.

 

[mathmari]

$|G/N| = q$ (all groups of order $q$ are abelian, in fact, they are cyclic, and simple).

Why are all groups of order $q$ are abelian? (Wondering)
I got stuck right now...

 

[Deveno]

Why are all groups of order $q$ are abelian? (Wondering)
I got stuck right now...

Because $q$ is a prime...

 

[mathmari]

$|G/N| = q$ (all groups of order $q$ are abelian, in fact, they are cyclic, and simple).

$|G/N| = p^2$ (all groups of order $p^2$ are abelian...why?)

We have that when $|G|=p^n$ for some $n$ then $Z(G)\neq 1$.

A group $G$ of order $|G|=q$, where $q$ is a prime is abelian:
We have that $Z(G)\neq \{1\}$, because of the above sentence.
So $|Z(G)|\neq 1$.
Since $|Z(G)|\mid |G|$, we must have that $|Z(G)|=|G|$.
Does this mean that $G=Z(G)$ ? (Wondering)
If that is true, it implies that $G$ is abelian, right? (Wondering) A group $G$ of order $|G|=p^2$, where $p$ is a prime is abelian:
We have that $Z(G)\neq \{1\}$, because of the above sentence.
So $|Z(G)|\neq 1$.
Since $|Z(G)|\mid |G|$, we must have that $|Z(G)|\in \{p, p^2\}$.
If $|Z(G)|=p^2$, then $G=Z(G)$, so $G$ is abelian, right? (Wondering)
If $|Z(G)|=p$, then $|G/Z(G)|=\frac{|G|}{|Z(G)|}=\frac{p^2}{p}=p$. So $G/Z(G)$ is cyclic. Does this imply that $G$ is abelian? (Wondering)

 

[Deveno]

We have that when $|G|=p^n$ for some $n$ then $Z(G)\neq 1$.

A group $G$ of order $|G|=q$, where $q$ is a prime is abelian:
We have that $Z(G)\neq \{1\}$, because of the above sentence.
So $|Z(G)|\neq 1$.
Since $|Z(G)|\mid |G|$, we must have that $|Z(G)|=|G|$.
Does this mean that $G=Z(G)$ ? (Wondering)
If that is true, it implies that $G$ is abelian, right? (Wondering)

It's even simpler that that: if $q$ is a prime, and $|G| = q$, then if $x \neq e$ is a non-identity element of $G$, the order of $x$ must be $q$ (since the order of $x$ divides $q$, and the order of $x$ is not 1, since $x \neq e$). Hence $x$ generates $G$, that is, $G$ is cyclic of order $q$. Of course, every cyclic group is abelian.

A group $G$ of order $|G|=p^2$, where $p$ is a prime is abelian:
We have that $Z(G)\neq \{1\}$, because of the above sentence.
So $|Z(G)|\neq 1$.
Since $|Z(G)|\mid |G|$, we must have that $|Z(G)|\in \{p, p^2\}$.
If $|Z(G)|=p^2$, then $G=Z(G)$, so $G$ is abelian, right? (Wondering)
If $|Z(G)|=p$, then $|G/Z(G)|=\frac{|G|}{|Z(G)|}=\frac{p^2}{p}=p$. So $G/Z(G)$ is cyclic. Does this imply that $G$ is abelian? (Wondering)

Yes, it does (sort of, because it can't happen that way)...do you know why?

 

[mathmari]

It's even simpler that that: if $q$ is a prime, and $|G| = q$, then if $x \neq e$ is a non-identity element of $G$, the order of $x$ must be $q$ (since the order of $x$ divides $q$, and the order of $x$ is not 1, since $x \neq e$). Hence $x$ generates $G$, that is, $G$ is cyclic of order $q$. Of course, every cyclic group is abelian.

Ah ok... I see... (Thinking)

Yes, it does (sort of, because it can't happen that way)...do you know why?

Not really... Could you explain it to me? (Wondering)

 

[Deveno]

Theorem: If $G/Z(G)$ is cyclic, $G$ is abelian.

Suppose $G/Z(G) = \langle xZ(G)\rangle$.

Since the cosets of $Z(G)$ partition $G$, we can write $g_1,g_2 \in G$ as:

$g_1 = x^kz_1$, for some $k$ and $z_1 \in Z(G)$ and
$g_2 = x^mz_2$, for some $m$ and $z_2 \in Z(G)$.

Then $g_1g_2 = (x^kz_1)(x^mz_2) = (x^kz_1)(z_2x^m) = x^k(z_1z_2)x^m$

$= x^kx^m(z_1z_2)$ (since $Z(G)$ is a subgroup, and thus $z_1z_2 \in Z(G)$ and commutes with all of $G$)

$= x^{k+m}(z_1z_2) = x^{m+k}(z_1z_2) = (x^mx^k)(z_1z_2)$

$= x^m(x^kz_1)z_2 = x^m(z_2(x^kz_1)) = (x^mz_2)(x^kz_1) = g_2g_1$,

so $G$ is abelian. (And then $Z(G) = G$, so $G/Z(G)$ is the cyclic trivial group of order 1).