How many Sylow-$ 3$ subgroup does $G$ have?

Question

Let $G$ be a noncyclic group of order $21$. How many sylow-$3$ subgroup does G have?

The possible orders of Sylow $3$ subgroups is $1, 7$. But how to check the exact number?

Answer 1

If it is $1$, $G$ must be cyclic as Sylow-$7$ subgroup is uniqe so it must be $7$.

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Notice that $n_7$ must be equal to $1$, so it has a uniqe sylow-$7$ subgroup.

As you said $n_3\in \{1,7\}$, if $n_3=1$;

it has also normal subgroup of order $3$ which means $G=HK$ and $H\cap K=1$ and $H,K$ is normal in $G$ which means that $G$ is abelian.

Since order of $H,K$ is relativly prime, $G$ must be cyclic which is a contradiction.

Answer 2

Already you have noticed number of Sylow-3 subgroup=$n_3 \in \{1,7\}$ and note that Sylow-$7$ subgroup is unique,hence normal.

If now Sylow-$3$ subgroup is unique then say $H$ and $K$ are the Sylow-$3$ and Sylow-$7$ subgroups respectively.

Then clearly $H\cap K=\{e\}$ and thus $|HK|=|G|=21\Rightarrow G=HK$

but we see that $\forall h\in H,k\in K, hkh^{-1}k^{-1}\in H\cap K=\{e\}$ because $h(kh^{-1}k^{-1})\in H$ as $H$ is normal and similarly $(hkh^{-1})k^{-1}\in K$ as $K$ is normal.So $hkh^{-1}k^{-1}=e\Rightarrow hk=kh$. More over both $H$ and $K$ are cyclic and hence abelian.

Thus $G$ is abelian.

Now we see that as $G$ must have element of order $3$ and element of order $7$ (By Cauchy's Theorem) let $o(a)=3,o(b)=7$ Since G is abelian $ab=ba$ and also we see gcd $(3,7)=1$

Thus $o(ab)=3 \times 7=21=|G|\Rightarrow G$ cyclic which is a contradiction. Hence $n_3=7$

Answer 3

By the way, you can realize this group concretely in the following form. Let $F = \mathbb{Z}/7\mathbb{Z} = \{0, 1, \dots, 6 \}$ be the field with $7$ elements, and $$ G = \left\{ \begin{bmatrix} 1 & a\\0 & b\end{bmatrix} : a \in F, b \in \{1, 2, 4 \} \right\}. $$ Note that $\{1, 2, 4 \}$ is a subgroup of $F^{\star}$ of order $3$.

You can verify that $$ G = \left\{ \begin{bmatrix} 1 & a\\0 & 1\end{bmatrix} : a \in F \right\} $$ is the $7$-Sylow subgroup, and that all elements outside it have order $3$. This is because $$ \begin{bmatrix} 1 & a\\0 & b\end{bmatrix}^{3} = \begin{bmatrix} 1 & a(1+b+b^{2})\\0 & b^{3}\end{bmatrix}, $$ and for $b \in \{ 2, 4 \}$ we have $b^{2} + b + 1 = 0$ in $F$ (and clearly $b^{3} = 1$).