MHB Is Every Group of Order 135 Nilpotent?

[mathmari]

Hey!

I want to show that each group of order $135$ is nilpotent.

We have that $G$ is called nilpotent iff there is a series of normal subgroups $$1\leq N_1\leq N_2\leq \dots \leq N_k=G$$ such that $N_{i+1}/N_i\subseteq Z(G/N_i)\Leftrightarrow [G,N_{i+1}]\subseteq N_i$, right? (Wondering)

Could you maybe explain to me how we can find such a series? (Wondering)

 

[mathmari]

I thought about it again... (Thinking) Since the order the group $G$ is odd, we have from Feit-Thompson Theorem that $G$ is solvable.

From that it follows that $G$ is nilpotent.

Is this correct? (Wondering) Or do we have to use the definition of a nilpotent group to show that? (Wondering)

 

[Deveno]

I suspect you are supposed to use Sylow theory, here.

Some lemmas you should use along the way:

The direct product of two nilpotent groups is nilpotent.

$p$-groups are nilpotent.

It is not true that solvable groups are nilpotent, the implication goes the other way. For a counter-example, consider $S_3$. This is solvable, we have the normal series:

$1 \lhd A_3 \lhd S_3$

and $S_3/A_2 \cong C_2$, with $A_3/1 \cong C_3$.

However, there is no finite upper central series, since $S_3$ is centerless (its center is trivial, so the upper central series never terminates, it goes:

$1 \lhd 1 \lhd 1 \lhd 1 \dots$)

 

[mathmari]

I suspect you are supposed to use Sylow theory, here.

Some lemmas you should use along the way:

The direct product of two nilpotent groups is nilpotent.

$p$-groups are nilpotent.

So, we use the fact that a group of order $p^a$ is nilpotent, right? (Wondering)

We have that $135=3^3\cdot 5$.

A group of order $3^3$ is nilpotent and a group of order $5$ is also nilpotent.

So, is a group of order $135$ a product of a group of order $3^3$ and a group of order $5$ ? (Wondering)

If yes, since the product of two nilpotent groups is nilpotent, we conclude that a group of order $135$ is nilpotent, right? (Wondering)

It is not true that solvable groups are nilpotent, the implication goes the other way. For a counter-example, consider $S_3$. This is solvable, we have the normal series:

$1 \lhd A_3 \lhd S_3$

and $S_3/A_2 \cong C_2$, with $A_3/1 \cong C_3$.

However, there is no finite upper central series, since $S_3$ is centerless (its center is trivial, so the upper central series never terminates, it goes:

$1 \lhd 1 \lhd 1 \lhd 1 \dots$)

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

 

[Deveno]

So, we use the fact that a group of order $p^a$ is nilpotent, right? (Wondering)

Yes...if you've proven that already.

We have that $135=3^3\cdot 5$.

A group of order $3^3$ is nilpotent and a group of order $5$ is also nilpotent.

So, is a group of order $135$ a product of a group of order $3^3$ and a group of order $5$ ? (Wondering)

You'd have to first show that we have a direct product (what property of the Sylow subgroups must we show for that)?

If yes, since the product of two nilpotent groups is nilpotent, we conclude that a group of order $135$ is nilpotent, right? (Wondering)

It's important that it be a *direct* product.

 

[mathmari]

Yes...if you've proven that already.

Yes, in my notes there is the following proposition:

Every finite $p$-group is nilpotent.

It's important that it be a *direct* product.

What do you mean by a direct product? (Wondering)

 

[Deveno]

There are 4 possible ways "product" of subgroups $H,K$ could be interpreted:

1. The join $H \vee K = \langle H,K\rangle = \langle H \cup K\rangle$

2. The product set $HK = \{hk: h \in H, K \in K\}$, which is a group if and only iff $HK = KH$

3. The (internal) semi-direct product $G=HK$, where $H \cap K = \{1\}$ and one of $H,K$ is normal.

4. The (internal) direct product $G = HK$, where $H \cap K = \{1\}$ and both of $H,K$ are normal.

Each category is increasingly restrictive, either on normality of the subgroups, how they interact, or their intersection. We're looking for category 4, the most restrictive condition.

 

[mathmari]

The (internal) direct product $G = HK$, where $H \cap K = \{1\}$ and both of $H,K$ are normal.

We have that $|G|=3^3\cdot 5$.

So, there are $3$-Sylow and $5$-Sylow in $G$.

Do we have to show that there is only one of each of them to show that they are normal? (Wondering)

 

[Deveno]

We have that $|G|=3^3\cdot 5$.

So, there are $3$-Sylow and $5$-Sylow in $G$.

Do we have to show that there is only one of each of them to show that they are normal? (Wondering)

Yes.

 

[mathmari]

We have that $|G|=3^3\cdot 5$.

So, there are $3$-Sylow and $5$-Sylow in $G$.

$$P\in \text{Syl}_3(G), \ |P|=3^3 \\ Q\in \text{Syl}_5(G), \ |Q|=5$$

We have that $n_3=1+3k$ and $n_3\mid 3^3\cdot 5 \Rightarrow n_3\mid 5$.

So, $n_3=1$.

We have that $n_5=1+5n$ and $n_5\mid 3^3\cdot 5 \Rightarrow n_5\mid 3^3$.

So, $n_5=1$.

Therefore, the subgroups $P$ and $Q$ are normal in $G$, right? (Wondering) Now it is left to show that $P\cap Q=\{1\}$, or not? (Wondering)

We have that $P$ has $3^3-1=26$ elements of order $3^3$ and $Q$ has $5-1=4$ elements of order $5$ and at the intersection there is just the identity.
Is this correct? (Wondering) Therefore, we have that $G=PQ$.

Since $P$ and $Q$ are $p$-groups, they are nilpotent.
So, $G=PQ$ is also nilpotent, right? (Wondering)

 

[Deveno]

We have that $|G|=3^3\cdot 5$.

So, there are $3$-Sylow and $5$-Sylow in $G$.

$$P\in \text{Syl}_3(G), \ |P|=3^3 \\ Q\in \text{Syl}_5(G), \ |Q|=5$$

We have that $n_3=1+3k$ and $n_3\mid 3^3\cdot 5 \Rightarrow n_3\mid 5$.

So, $n_3=1$.

We have that $n_5=1+5n$ and $n_5\mid 3^3\cdot 5 \Rightarrow n_5\mid 3^3$.

So, $n_5=1$.

Therefore, the subgroups $P$ and $Q$ are normal in $G$, right? (Wondering) Now it is left to show that $P\cap Q=\{1\}$, or not? (Wondering)

We have that $P$ has $3^3-1=26$ elements of order $3^3$ and $Q$ has $5-1=4$ elements of order $5$ and at the intersection there is just the identity.
Is this correct? (Wondering) Therefore, we have that $G=PQ$.

Since $P$ and $Q$ are $p$-groups, they are nilpotent.
So, $G=PQ$ is also nilpotent, right? (Wondering)

Yes, that is correct. A few comments:

1. It is obvious that $P \cap Q = \{1\}$, since they belong to distinct primes.

2. There is a difference between saying $G = PQ$ and $G \cong P \times Q$, to show nilpotency, we want the latter. This is actually an "iff" theorem:

$G$ is nilpotent if and only if it is the direct product of its Sylow subgroups.

Fortunately, you only have to use the "if" part, here. We can discuss the "only if" part, if you like, later.

One thing sort of "missing" (you have not said you have a proof of it) is the proof of the statement:

The direct product of two nilpotent groups, is nilpotent. This rests on another theorem/lemma:

If $G = \prod\limits_k G_k$, then the $i$-th center of $G$ is the direct product of the $i$-th centers of the $G_k$:

$Z_i(G) = \prod\limits_k Z_i(G_k)$

or, as stated for just two factors:

If $G = H \times K$, then $Z_i(G) = Z_i(H) \times Z_i(K)$

 

[mathmari]

There is a difference between saying $G = PQ$ and $G \cong P \times Q$, to show nilpotency, we want the latter. This is actually an "iff" theorem:

$G$ is nilpotent if and only if it is the direct product of its Sylow subgroups.

I found in my notes the following proposition:

If $N,M\trianglelefteq G$, where $N,M$ nilpotent, then $NM$ nilpotent. Can we not use this one? Or isn't this formulation correct? (Wondering)
In my notes there is also a theorem according to which the following two statements are equivalent:

• $G$ is nilpotent.
• All its Sylow subgroups are normal.

We have shown that $n_3=n_5=1$, so the $3$-Sylow and $5$-Sylow subgroups are normal. Can we conclude also in that way that $G$ is nilpotent? (Wondering)

 

[Deveno]

I found in my notes the following proposition:

If $N,M\trianglelefteq G$, where $N,M$ nilpotent, then $NM$ nilpotent. Can we not use this one? Or isn't this formulation correct? (Wondering)
In my notes there is also a theorem according to which the following two statements are equivalent:

• $G$ is nilpotent.
• All its Sylow subgroups are normal.

We have shown that $n_3=n_5=1$, so the $3$-Sylow and $5$-Sylow subgroups are normal. Can we conclude also in that way that $G$ is nilpotent? (Wondering)

Yes, that is the "iff" I referred to. We have shown any group of order 135 has a normal Sylow 3-subgroup, and a normal Sylow 5-subgroup, and since $G$ is *always* the product set of Sylow $p_i$-subgroups where the $p_i$ are distinct primes dividing the order of $G$ (we can argue by order, here), we have:

$G = PQ$

and $P,Q \lhd G$.

It is trivial that Sylow subgroups belonging to distinct primes have only the identity in their intersection, for example, for a Sylow $p$-subgroup $P$, with $x \in P$, the order of $x$ is a power of $p$, and similarly for $y \in Q$ a Sylow $q$-subgroup.

The only power of $p$ that is also a power of $q$ is $p^0 = q^0 = 1$, so all (the only) elements of the intersection have order 1.

The upshot of this, is that:

All Sylow subgroups of $G$ are normal $\iff$ $G$ is the direct product of its Sylow subgroups.

One caveat, here: make sure you understand WHY $N,M$ normal and each nilpotent implies $NM$ is nilpotent. It's OK to use material from a text, or notes, but if you don't follow the proof, then you have "holes" in your understanding of the theorems you "build from them".

 

[mathmari]

I got stuck right now... Now we have that $G=PQ$, right? Do we have to show that $G \cong P \times Q$ ? (Wondering)

 

[Deveno]

I got stuck right now... Now we have that $G=PQ$, right? Do we have to show that $G \cong P \times Q$ ? (Wondering)

For ANY group with order divisible by just two primes, say $p$ and $q$, we *always* have $G = PQ$. That does not mean the Sylow subgroups are necessarily normal, which is the difference between $G = PQ$ and $G = P \times Q$.

IN THIS CASE, we have shown that, so we're good.