Can a group of order $2^6\cdot 5^6$ be simple?

[mathmari]

Hey!

I want to show that if $|G|=pqr$ where $p,g,r$ are primes, then $G$ is not simple.

We have that a group is simple if it doesn't have any non-trivial normal subgroups, right? (Wondering)

Could you give me some hints how we could show that the above group is not simple? (Wondering)

Do we have to suppose that $G$ is simple? Then it doesn't have any non-trivial normal subgroups.

But how do we get a contradiction? (Wondering)

 

[mathmari]

There are $p$-Sylow, $q$-Sylow and $r$-Sylow in $G$.

$$P\in \text{Syl}_p(G), \ |P|=p, \\ Q\in \text{Syl}_q(G), \ |Q|=q, \\ R\in \text{Syl}_r(G), \ |R|=r$$

$|\text{Syl}_p(G)|=1+kp \mid |G|=pqr$ so $1+kp\mid pqr \Rightarrow 1+kp\mid qr$.

So, $$1+kp=\left\{\begin{matrix}
1 \\
q \\
r \\
qr
\end{matrix}\right.$$ Similarily, $|\text{Syl}_q(G)|=1+kq \mid |G|=pqr$ so $1+kq\mid pqr \Rightarrow 1+kq\mid pr$.

So, $$1+kq=\left\{\begin{matrix}
1 \\
p \\
r \\
pr
\end{matrix}\right.$$ And, $|\text{Syl}_r(G)|=1+kr \mid |G|=pqr$ so $1+kr\mid pqr \Rightarrow 1+kr\mid pq$.

So, $$1+kr=\left\{\begin{matrix}
1 \\
p \\
q \\
pq
\end{matrix}\right.$$

Is this correct? (Wondering) Suppose that $|\text{Syl}_p(G)|=1$ or $|\text{Syl}_q(G)|=1$ or $|\text{Syl}_r(G)|=1$. That means that there is one Sylow subgroup that has order $p$, $q$, $r$, respectively, right? Is this Sylow subgroup normal? (Wondering)

Suppose that $|\text{Syl}_p(G)|=q$. That means that the number of $p$-Sylow subgroups is $q$, each of which has order $p$, right? How can we check if these subgroups are normal? (Wondering)

 

[Deveno]

Try finding the number of Sylow subgroups of the LARGEST prime of $p,q,r$. Suppose it's $r$.

By your own argument, we must have either $1+ kr = 1$, or $1 + kr = pq$ (We cannot have $1+kr = p$ or $1 + kr = q$ since $p,q < r$).

And yes, if we have just ONE Sylow subgroup of a given prime, it is normal, since Sylow subgroups are conjugate.

Now assume $G$ is NOT simple, and count the number of elements of order $p,q$ and $r$.

Note this assumes $p < q < r$, and it may be that $p = r$ (that is, our group has order $p^2q$ for $p < q$). This is another case that needs to be handled separately.

Finally, if $p = q = r$, we have a group of order $p^3$. There is a theorem that says any $p$-group has a non-trivial center, which gives us a non-trivial proper normal subgroup in the non-abelian case. In the abelian case, we can take any subgroup of order $p$ (which exists by the Sylow theorems, or Cauchy's Theorem).

 

[mathmari]

Try finding the number of Sylow subgroups of the LARGEST prime of $p,q,r$. Suppose it's $r$.

By your own argument, we must have either $1+ kr = 1$, or $1 + kr = pq$ (We cannot have $1+kr = p$ or $1 + kr = q$ since $p,q < r$).

And yes, if we have just ONE Sylow subgroup of a given prime, it is normal, since Sylow subgroups are conjugate.

Now assume $G$ is NOT simple, and count the number of elements of order $p,q$ and $r$.

Note this assumes $p < q < r$

There are $p$-Sylow, $q$-Sylow and $r$-Sylow in $G$.

$$P\in \text{Syl}_p(G), \ |P|=p, \\ Q\in \text{Syl}_q(G), \ |Q|=q, \\ R\in \text{Syl}_r(G), \ |R|=r$$

$|\text{Syl}_p(G)|=1+kp \mid |G|=pqr$ so $1+kp\mid pqr \Rightarrow 1+kp\mid qr$.

So, $$1+kp=\left\{\begin{matrix}
1 \\
q \\
r \\
qr
\end{matrix}\right.$$ Similarily, $|\text{Syl}_q(G)|=1+kq \mid |G|=pqr$ so $1+kq\mid pqr \Rightarrow 1+kq\mid pr$.

So, $$1+kq=\left\{\begin{matrix}
1 \\
p \\
r \\
pr
\end{matrix}\right.$$ And, $|\text{Syl}_r(G)|=1+kr \mid |G|=pqr$ so $1+kr\mid pqr \Rightarrow 1+kr\mid pq$.

So, $$1+kr=\left\{\begin{matrix}
1 \\
p \\
q \\
pq
\end{matrix}\right.$$
If $|\text{Syl}_p(G)|=1$ or $|\text{Syl}_q(G)|=1$ or $|\text{Syl}_r(G)|=1$ then the corresponding Sylow is unique and therefore normal in $G$, so $G$ not simple.

We suppose that the primes $p, q, r$ are different, let $p<q<r$.
If $|\text{Syl}_r(G)|\neq 1$, then $|\text{Syl}_r(G)|=pq$, because it cannot hold that $1+kr = p$ or $1 + kr = q$ since $p,q < r$.

Since $|\text{Syl}_r(G)|=pq$ we have $pq$ $r$-Sylow subgroups, each of which has order $r$.
They have as intersection the "$1$".
So, in total the number of elements of order $r$ in all $pq$ above subgroups is $pq(r-1)=pqr-pq$.
Suppose that $|\text{Syl}_q(G)|\neq 1$ then since there are $|\text{Syl}_q(G)|=1+kq>q>p$ $q$-Sylow subgroups, we have at least $p(q-1)=pq-p$ elements of order $q$.
Therefore, in total we have at least $pqr-pq+pq-p=pqr-p$ elements of order $r$ or $q$.
Since $|G|=pqr$, we can have only one unique $p$-Sylow subgroups, that is then normal.

Therefore, $G$ is not simple.
Is everything correct? (Wondering)

 

[Deveno]

There are $p$-Sylow, $q$-Sylow and $r$-Sylow in $G$.

$$P\in \text{Syl}_p(G), \ |P|=p, \\ Q\in \text{Syl}_q(G), \ |Q|=q, \\ R\in \text{Syl}_r(G), \ |R|=r$$

$|\text{Syl}_p(G)|=1+kp \mid |G|=pqr$ so $1+kp\mid pqr \Rightarrow 1+kp\mid qr$.

So, $$1+kp=\left\{\begin{matrix}
1 \\
q \\
r \\
qr
\end{matrix}\right.$$ Similarily, $|\text{Syl}_q(G)|=1+kq \mid |G|=pqr$ so $1+kq\mid pqr \Rightarrow 1+kq\mid pr$.

So, $$1+kq=\left\{\begin{matrix}
1 \\
p \\
r \\
pr
\end{matrix}\right.$$ And, $|\text{Syl}_r(G)|=1+kr \mid |G|=pqr$ so $1+kr\mid pqr \Rightarrow 1+kr\mid pq$.

So, $$1+kr=\left\{\begin{matrix}
1 \\
p \\
q \\
pq
\end{matrix}\right.$$
If $|\text{Syl}_p(G)|=1$ or $|\text{Syl}_q(G)|=1$ or $|\text{Syl}_r(G)|=1$ then the corresponding Sylow is unique and therefore normal in $G$, so $G$ not simple.

We suppose that the primes $p, q, r$ are different, let $p<q<r$.
If $|\text{Syl}_r(G)|\neq 1$, then $|\text{Syl}_r(G)|=pq$, because it cannot hold that $1+kr = p$ or $1 + kr = q$ since $p,q < r$.

Since $|\text{Syl}_r(G)|=pq$ we have $pq$ $r$-Sylow subgroups, each of which has order $r$.
They have as intersection the "$1$".
So, in total the number of elements of order $r$ in all $pq$ above subgroups is $pq(r-1)=pqr-pq$.
Suppose that $|\text{Syl}_q(G)|\neq 1$ then since there are $|\text{Syl}_q(G)|=1+kq>q>p$ $q$-Sylow subgroups, we have at least $p(q-1)=pq-p$ elements of order $q$.
Therefore, in total we have at least $pqr-pq+pq-p=pqr-p$ elements of order $r$ or $q$.
Since $|G|=pqr$, we can have only one unique $p$-Sylow subgroups, that is then normal.

Therefore, $G$ is not simple.
Is everything correct? (Wondering)

You've got the right idea, but since $n_q > 1$ (that is $k \neq 0$, so $k \geq 1$), we have $n_q = 1 + kq > q > p$.

Thus the number of elements of order $q$ is strictly greater than $pq - p$.

Hence the number of elements of $G$ of order $q$ or $r$ is strictly greater than $pqr - p$.

Since we have one element of order 1, and at least $p-1$ elements of order $p$, we have:

$|G| > (pqr - p) + (p-1) + 1 = pqr = |G|$, a contradiction, so $G$ must not be simple.

(Here, I am using $n_q$ for $|\text{Syl}_q(G)|$, for brevity's sake).

As I mentioned before, this handles the case where $p,q,r$ are DISTINCT primes, but the original problem you posed does not say this, so there are other cases you may have to consider.

 

[mathmari]

Since we have one element of order 1, and at least $p-1$ elements of order $p$, we have:

$|G| > (pqr - p) + (p-1) + 1 = pqr = |G|$, a contradiction, so $G$ must not be simple.

Could you explain to me why we have one element of order $1$ and at least $p-1$ elements of order $p$ ? (Wondering)

Do we suppose at the beginning at $G$ is simple? Or for what do we get a contradiction? (Wondering)

 

[Deveno]

Could you explain to me why we have one element of order $1$ and at least $p-1$ elements of order $p$ ? (Wondering)

Do we suppose at the beginning at $G$ is simple? Or for what do we get a contradiction? (Wondering)

Because $p$ divides the order of $G$, and so $G$ has an element of order $p$ which generates a subgroup of $G$ of order $p$, and any group of order $p$ has $p-1$ elements of order $p$, the identity, of curse, has order 1.

We used the assumption that $G$ was not simple to force an "overcount" of the elements of $G$.

(EDIT: oops! see below).

 

[mathmari]

Because $p$ divides the order of $G$, and so $G$ has an element of order $p$ which generates a subgroup of $G$ of order $p$, and any group of order $p$ has $p-1$ elements of order $p$, the identity, of curse, has order
1.

Ah ok... But why do we say that we have at least $p-1$ elements of order $p$ ? (Wondering)

We used the assumption that $G$ was not simple to force an "overcount" of the elements of $G$.

When we assume that $G$ is not simple and we get a contradiction, it follows that $G$ is simple, or not? (Wondering)

 

[Deveno]

Ah ok... But why do we say that we have at least $p-1$ elements of order $p$ ? (Wondering)

The $p$-Sylow subgroups have order $p$, we have $1+kp$ of them, and this number cannot be 0. Basically, we are using simple counting to show that $pq$ subgroups of order $r$ is "too many".

When we assume that $G$ is not simple and we get a contradiction, it follows that $G$ is simple, or not? (Wondering)

I said that wrong-we assumed $G$ was simple. Sorry for the confusion.

 

[mathmari]

The $p$-Sylow subgroups have order $p$, we have $1+kp$ of them, and this number cannot be 0.

So, we know that $|\text{Syl}_p(G)|\geq 1$.
Since a $p$-Sylow subgroup has order $p$, and so $p-1$ elements have order $p$, we have that in total the number of the elements of all the $p$-Sylow subgroups that have order $p$ is greater or equal to $p-1$, right? (Wondering)

I said that wrong-we assumed $G$ was simple. Sorry for the confusion.

At which point do we use the assumption that $G$ is simple? (Wondering)

 

[Deveno]

So, we know that $|\text{Syl}_p(G)|\geq 1$.
Since a $p$-Sylow subgroup has order $p$, and so $p-1$ elements have order $p$, we have that in total the number of the elements of all the $p$-Sylow subgroups that have order $p$ is greater or equal to $p-1$, right? (Wondering)

Sure thing.

At which point do we use the assumption that $G$ is simple? (Wondering)

When we assume that $|\text{Syl}_r(G)| > 1$, (If it was 1, then the Sylow $r$-subgroup would be normal, and thus be a non-trivial normal subgroup). We also use it later when we assume likewise that $|\text{Syl}_q(G)| > q$ (that is we assume that it is greater than one, so it is at the very least, greater than or equal to $1 + q$, which is greater than $q$, which is greater than $p$).

Each time we assume we have more than 1 Sylow subgroup for any of $p,q,r$, we are using the assumed simplicity of $G$. This eventually leads to a contradiction, because we wind up with too many elements.

 

[mathmari]

When we assume that $|\text{Syl}_r(G)| > 1$, (If it was 1, then the Sylow $r$-subgroup would be normal, and thus be a non-trivial normal subgroup). We also use it later when we assume likewise that $|\text{Syl}_q(G)| > q$ (that is we assume that it is greater than one, so it is at the very least, greater than or equal to $1 + q$, which is greater than $q$, which is greater than $p$).

Each time we assume we have more than 1 Sylow subgroup for any of $p,q,r$, we are using the assumed simplicity of $G$. This eventually leads to a contradiction, because we wind up with too many elements.

It stands that if a Sylow subgroup is unique then it is normal, so the group is not simple, right? (Wondering)

We suppose that the primes $p, q, r$ are different, let $p<q<r$.
If $|\text{Syl}_r(G)|\neq 1$, then $|\text{Syl}_r(G)|=pq$, because it cannot hold that $1+kr = p$ or $1 + kr = q$ since $p,q < r$.

Since $|\text{Syl}_r(G)|=pq$ we have $pq$ $r$-Sylow subgroups, each of which has order $r$.
They have as intersection the "$1$".
So, in total the number of elements of order $r$ in all $pq$ above subgroups is $pq(r-1)=pqr-pq$.
Suppose that $|\text{Syl}_q(G)|\neq 1$ then since there are $|\text{Syl}_q(G)|=1+kq>q>p$ $q$-Sylow subgroups, we have at least $p(q-1)=pq-p$ elements of order $q$.
Therefore, in total we have at least $pqr-pq+pq-p=pqr-p$ elements of order $r$ or $q$.
Since $|G|=pqr$, we can have only one unique $p$-Sylow subgroups, that is then normal.

Therefore, $G$ is not simple.

I got the basic idea of an other example of my notes... But I haven't really understood why the intersection of the $pq$ $r$-Sylow subgroups, each of which has order $r$, is "$1$"... Could you explain it to me? (Wondering)

 

[Deveno]

It stands that if a Sylow subgroup is unique then it is normal, so the group is not simple, right? (Wondering)
I got the basic idea of an other example of my notes... But I haven't really understood why the intersection of the $pq$ $r$-Sylow subgroups, each of which has order $r$, is "$1$"... Could you explain it to me? (Wondering)

These groups have order $r$, and $r$ is prime. Suppose we call two such subgroups $R_1$ and $R_2$. $R_1 \cap R_2$ is a subgroup of both $R_1$ and $R_2$, and so it has order dividing $r$. How many divisors does a prime have?

 

[mathmari]

These groups have order $r$, and $r$ is prime. Suppose we call two such subgroups $R_1$ and $R_2$. $R_1 \cap R_2$ is a subgroup of both $R_1$ and $R_2$, and so it has order dividing $r$. How many divisors does a prime have?

The order of the intersection is either $1$ or $r$, right? How can we reject that it is $r$ ? (Wondering)

 

[Deveno]

The order of the intersection is either $1$ or $r$, right? How can we reject that it is $r$ ? (Wondering)

If two subgroups of order $r$ have intersection of order $r$, they are the same subgroup, right?

 

[mathmari]

If two subgroups of order $r$ have intersection of order $r$, they are the same subgroup, right?

Yes. That cannot hold because then we would have that $|\text{Syl}_r(G)|=pq-1$, or not? (Wondering)

 

[Deveno]

Yes. That cannot hold because then we would have that $|\text{Syl}_r(G)|=pq-1$, or not? (Wondering)

Any two DIFFERENT Sylow $r$-subgroups must have trivial intersection. Since we have $pq$ of them (by assumption that $G$ is simple), none of them can share any members but the identity, giving us $pq(r - 1)$ elements of order $r$ in all.

The situation becomes more complicated when you consider subgroups of order $r^2$, because it is possible to have an intersection of order $r$, making it difficult to get an exact count of elements of order $r$. Fortunately, since $r^2 \not\mid |G|$, we don't have to worry about that in this example.

 

[mathmari]

Ah ok... I see... (Nod) I tried to apply this proof at a specific example... I want to show that a group of order $280$ is not simple.

We have that $|G|=280=2^3\cdot 5\cdot 7$. There are $2$-Sylow, $5$-Sylow and $7$-Sylow in $G$.

$$P\in \text{Syl}_2(G), \ |P|=2, \\ Q\in \text{Syl}_5(G), \ |Q|=5, \\ R\in \text{Syl}_7(G), \ |R|=7$$

$|\text{Syl}_2(G)|=1+2k \mid |G|=2^3\cdot 5\cdot 7$ so $1+2k\mid 2^3\cdot 5\cdot 7 \Rightarrow 1+2k\mid 5\cdot 7$.

So, $$|\text{Syl}_2(G)|=\left\{\begin{matrix}
1\\
5 \\
7 \\
5\cdot 7=35
\end{matrix}\right.$$ Similarily, $|\text{Syl}_5(G)|=1+5m \mid |G|=2^3\cdot 5\cdot 7$ so $1+5m\mid 2^3\cdot 5\cdot 7 \Rightarrow 1+5m\mid 2^3\cdot 7$.

So, $$1+5m=\left\{\begin{matrix}
1 \\
2 \\
2^2=4 \\
2^3=8 \\
7\\
2\cdot 7=14 \\
2^2\cdot 7=28 \\
2^3\cdot 7=56 \\
\end{matrix}\right.$$ And, $|\text{Syl}_7(G)|=1+7n \mid |G|=2^3\cdot 5\cdot 7$ so $1+7n\mid 2^3\cdot 5\cdot 7 \Rightarrow 1+7n\mid 2^3\cdot 5$.

So, $$1+7n=\left\{\begin{matrix}
1 \\
2 \\
2^2=4 \\
2^3=8 \\
5\\
2\cdot 5=10 \\
2^2\cdot 5=20 \\
2^3\cdot 5=40
\end{matrix}\right.$$
If $|\text{Syl}_2(G)|=1$ or $|\text{Syl}_5(G)|=1$ or $|\text{Syl}_7(G)|=1$ then the corresponding Sylow is unique and therefore normal in $G$, so $G$ not simple. Let $G$ be simple, that means that it must hold that $|\text{Syl}_7(G)|\neq 1$, then $|\text{Syl}_7(G)|=8$ or $|\text{Syl}_7(G)|=10$ or $|\text{Syl}_7(G)|=20$ or $|\text{Syl}_7(G)|=40$, because it cannot hold that $1+7n = 2$ or $1 + 7n = 4$ or $1+7n=5$ since $2, 4, 5 < 7$.
Is it correct so far? (Wondering)

Which $|\text{Syl}_7(G)|$ do we choose? The largest one? (Wondering)

 

[Deveno]

This is an entirely different kind of problem, the order of $G$ is not $pqr$, but $p^3qr$, which makes the analysis much more difficult.

However, we can argue this way:

If $G$ is simple, there cannot be just one Sylow 5-subgroup. But of the divisors of 56 that are greater than one, only 56 is congruent to 1 modulo 5. This gives 224 elements of order 5.

Similarly, of the divisors of 40 that are greater than 1, only 8 is congruent to 1 modulo 7. This gives an addition 48 elements of order 7, for a total of 272 elements.

But that only leaves 8 elements left over, so there is only room for 1 Sylow 2-subgroup, which is thus normal, contradicting our assumption $G$ is simple.

 

[mathmari]

But that only leaves 8 elements left over, so there is only room for 1 Sylow 2-subgroup, which is thus normal, contradicting our assumption $G$ is simple.

We have that $|\text{Syl}_2(G)|=\left\{\begin{matrix}
1\\
5 \\
7 \\
5\cdot 7=35
\end{matrix}\right.$

So, when $|\text{Syl}_2(G)|=5$ or $|\text{Syl}_2(G)|=7$, there are less than $8$ elements of order $2$, or not? (Wondering)

 

[Deveno]

We have that $|\text{Syl}_2(G)|=\left\{\begin{matrix}
1\\
5 \\
7 \\
5\cdot 7=35
\end{matrix}\right.$

So, when $|\text{Syl}_2(G)|=5$ or $|\text{Syl}_2(G)|=7$, there are less than $8$ elements of order $2$, or not? (Wondering)

Hard to say. All we know is that a Sylow 2-subgroup has 8 elements, we do not know from this information alone which group of order 8 it may be isomorphic to, it might be (isomorphic to):

$\Bbb Z_8$ (this has a single element of order 2)
$\Bbb Z_4 \times \Bbb Z_2$ (this has three elements of order 2)
$\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ (this has seven elements of order 2)
$D_4$ (this has five elements of order 2)
$Q_8$ (this has one element of order 2)

Any two of the Sylow 2-subgroups might intersect trivially, or in a group of order 2, or a group of order 4. This makes it difficult to say how many elements of order 2 we have overall.

There is a FUNDAMENTAL difference between a group of order $p$, and a $p$-group (a group whose order is a POWER of $p$). A group of order $p$ is necessarily cyclic, and simple, because it HAS no non-trivial proper subgroups at all, much less normal ones. This is not true (necessarily) of groups of order $p^k$, because they can come in many different "flavors".

Note I did NOT claim we have 8 elements of order 2, or even 7 elements of order 2. I only claimed that after counting the elements of order 7 and 5, we only have 8 elements left over, and this only leaves room for ONE subgroup of order 8 (no element of a group of order 8 can have order 5 or 7).

 

[mathmari]

Hard to say. All we know is that a Sylow 2-subgroup has 8 elements, we do not know from this information alone which group of order 8 it may be isomorphic to, it might be (isomorphic to):

$\Bbb Z_8$ (this has a single element of order 2)
$\Bbb Z_4 \times \Bbb Z_2$ (this has three elements of order 2)
$\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ (this has seven elements of order 2)
$D_4$ (this has five elements of order 2)
$Q_8$ (this has one element of order 2)

Any two of the Sylow 2-subgroups might intersect trivially, or in a group of order 2, or a group of order 4. This makes it difficult to say how many elements of order 2 we have overall.

There is a FUNDAMENTAL difference between a group of order $p$, and a $p$-group (a group whose order is a POWER of $p$). A group of order $p$ is necessarily cyclic, and simple, because it HAS no non-trivial proper subgroups at all, much less normal ones. This is not true (necessarily) of groups of order $p^k$, because they can come in many different "flavors".

Note I did NOT claim we have 8 elements of order 2, or even 7 elements of order 2. I only claimed that after counting the elements of order 7 and 5, we only have 8 elements left over, and this only leaves room for ONE subgroup of order 8 (no element of a group of order 8 can have order 5 or 7).

Ahh I got it! (Thinking)
I tried also an other example...

I want to show that a group of order $2^6\cdot 5^6$ is not simple.

I have done the following:

There are $2$-Sylow and $5$-Sylow in $G$.

$$P\in \text{Syl}_2(G), \ |P|=2^6, \\ Q\in \text{Syl}_5(G), \ |Q|=5^6$$

$|\text{Syl}_2(G)|=1+2k \mid |G|=2^6\cdot 5^6$ so $1+2k\mid 2^6\cdot 5^6 \Rightarrow 1+2k\mid 5^6$.

So, $$|\text{Syl}_2(G)|=\left\{\begin{matrix}
1\\
5 \\
5^2 \\
5^3 \\
5^4 \\
5^5 \\
5^6
\end{matrix}\right.$$ Similarily, $|\text{Syl}_5(G)|=1+5m \mid |G|=2^6\cdot 5^6$ so $1+5m\mid 2^6\cdot 5^6 \Rightarrow 1+5m\mid 2^6$.

So, $$1+5m=\left\{\begin{matrix}
1 \\
2 \\
2^4=16
\end{matrix}\right.$$ Let $G$ be simple.
So, there cannot be just one $5$-Sylow subgroup.
Therefore there are $16$ $5$-Syow subgroups of order $5^6$.
So, in total there are $16\cdot 5^6=2^4\cdot 5^6$ elements.

Since $|G|=2^6\cdot 5^6$ we can only have $$|G|-2^4\cdot 5^6=2^6\cdot 5^6-2^4\cdot 5^6=2^4\cdot 5^6(2^2-1)=2^4\cdot 5^6\cdot 3$$

Is it correct so far? (Wondering)

How could we continue? How could we conclude that we can only have one $2$-Sylow subgroup? (Wondering)

 

[mathmari]

I read again the first example and I have question...

Since we have one element of order 1, and at least $p-1$ elements of order $p$, we have:

The element of order $1$ is the element of the $p$-Sylow subgroup, or of all the Sylow subgroups? (Wondering)

 

[Deveno]

Ahh I got it! (Thinking)
I tried also an other example...

I want to show that a group of order $2^6\cdot 5^6$ is not simple.

I have done the following:

There are $2$-Sylow and $5$-Sylow in $G$.

$$P\in \text{Syl}_2(G), \ |P|=2^6, \\ Q\in \text{Syl}_5(G), \ |Q|=5^6$$

$|\text{Syl}_2(G)|=1+2k \mid |G|=2^6\cdot 5^6$ so $1+2k\mid 2^6\cdot 5^6 \Rightarrow 1+2k\mid 5^6$.

So, $$|\text{Syl}_2(G)|=\left\{\begin{matrix}
1\\
5 \\
5^2 \\
5^3 \\
5^4 \\
5^5 \\
5^6
\end{matrix}\right.$$ Similarily, $|\text{Syl}_5(G)|=1+5m \mid |G|=2^6\cdot 5^6$ so $1+5m\mid 2^6\cdot 5^6 \Rightarrow 1+5m\mid 2^6$.

So, $$1+5m=\left\{\begin{matrix}
1 \\
2 \\
2^4=16
\end{matrix}\right.$$ Let $G$ be simple.
So, there cannot be just one $5$-Sylow subgroup.
Therefore there are $16$ $5$-Syow subgroups of order $5^6$.
So, in total there are $16\cdot 5^6=2^4\cdot 5^6$ elements.

Since $|G|=2^6\cdot 5^6$ we can only have $$|G|-2^4\cdot 5^6=2^6\cdot 5^6-2^4\cdot 5^6=2^4\cdot 5^6(2^2-1)=2^4\cdot 5^6\cdot 3$$

Is it correct so far? (Wondering)

How could we continue? How could we conclude that we can only have one $2$-Sylow subgroup? (Wondering)

Again, the reasoning used in the $pqr$-case doesn't help us much here. It might be the case that the Sylow 5-subgroups all intersect in a group of order $5^5$, so that we can only count on there being:

$(5^6 - 5^5)16 + 5^5 = 5^6 - 5^5(15)$ elements in these subgroups.

That leaves $5^6(16) - 5^6(16) + 5^5(15) = 5^5(15)$ elements left over, which is enough to have $5^5$ Sylow 2-subgroups, even if they intersect trivially.

In a situation like this, it's better to use group actions. Suppose $P$ is a Sylow 5-subgroup. Then $P$ has index 16.

Letting $G$ act on the set of left cosets by left multiplication gives a homomorphism:

$\phi: G \to S_{16}$.

Can this homomorphism be injective? No, because $5^6$ divides the order of $G$, but $16!$ only has 3 factors of 5 (from 5, 10 and 15).

Can this homomorphism be trivial? No, because if $g \not\in P$, then $gP \neq P$, and so $g$ acts non-trivially on the coset space (that is, $g$ does not map to the identity of $S_{16}$).

We conclude that $\text{ker }\phi$ is not $G$, nor the identity of $G$, and is thus a non-trivial proper normal subgroup of $G$, which thus cannot be simple.

As regards your other question: in ANY group, there is always only ONE (and exactly one) element of order 1, for if:

$a^1 = e$, then $a = e$; that is the sole element of order one is always the identity.