Exercise 8.1.6

Proof.

(a)

Let $G$ be a group of order 10, and $N$ the number of 5-Sylow subgroups of $G$. Then $N\equiv 1(\mathit{mod}5)$ and $N\mid 10$. Therefore $N=1$. A group of order 10 has a unique 5-Sylow $H$, and as all 5-Sylow subgroup are conjugate, $H$ is normal in $G$, since the conjugate of a 5-Silow is a 5-Sylow. Then $|H|=5$ and $|G∕H|=2$ are prime, so $H$ and $G∕H$ are cyclic of prime order. This implies that $G$ is solvable.

Same reasoning if $|G|=15$: then $N\equiv 1(\mathit{mod}5)$, and $N\mid 15$, therefore $N=1$.

(b)

Let $G$ be a group of order 30.

$\text{∙}$

If $G$ is Abelian, a fortiori solvable, it contains a 5-Sylow subgroup, necessarily normal in $G$.

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If $G$ is a non Abelian solvable group, there exists a composition series $$\{e\}=G_n⊊G_{n-1}⊊\cdots ⊊G_0=G.$$

Then $n\ge 2$, otherwise a short series $\{e\}=G_1⊊G_0=G$, with $G_0∕G_1=G$ Abelian, is in contradiction with the hypothesis " $G$ non Abelian". Then the subgroup $G_1$ is normal in $G$, and $G_1\ne \{e\},G_1\ne G$.

Consequently the hypothesis $G$ solvable implies the existence of a proper non trivial subgroup of $G$.

$\text{∙}$

Conversely, suppose that $G$ has a normal subgroup $H$, with $H\ne \{e\},H\ne G$.

then $q=|H|$ divides 30, and $q\ne 1,q\ne 30$, so $q\in \{2,3,5,6,10,15\}$.

If $q=10$ or $q=15$, then $H$ is solvable by part (a), and the quotient group $G∕H$ is then of order $3$ or $2$ both prime, so $G∕H$ is cyclic. This proves that $G$ is solvable.

If $q=2$ or $q=3$, then $H$ is cyclic, and $G∕H$ of order 15 or 10 is solvable by part (a), so $G$ is solvable.

A group of order 5 is cyclic, a fortiori solvable, and a group of order 6 est isomorphic to the cyclic group $C_6$, or to $S_3$, and in both cases is solvable.

If $q=5$ or $q=6$, $H$ and $G∕H$ are both solvable, so $G$ is solvable.

Conclusion: a group $G$ of order 30 is solvable if and only if it contains a proper non trivial normal subgroup.

(c)

The number $N$ of 3-sylow subgroups of $G$ satisfies $N\equiv 1(\mathit{mod}3)$, and $N$ divides $30$, so $N=1$ or $N=10$.

The number $N^{\prime }$ of 5-sylow subgroups of $G$ satisfies $N\equiv 1(\mathit{mod}5)$, and $N^{\prime }$ divisdes $30$, so $N^{\prime }=1$ or $N^{\prime }=6$.

(d)

Suppose that $G$ contains ten 3-Sylow subgroups of $G$ and six 5-Sylow subgroups.

Two distinct cyclic subgroups of order $p$, with $p$ prime, have a trivial intersection, otherwise any non trivial element in the intersection would be a generator of these two subgroups, which would then be identical.

Consequently, two 3-Sylow subgroups have an intersection reduced to the identity of $G$, and this is the same for the 5-Sylow.

The union of 3-Sylow has then $1+10\times 2=21$ elements, and the union of the 5-Sylow $1+6\times 4=25$ elements. As a 5-Sylow and a 3-Sylow have a trivial intersection, $G$ would contains at least $21+25-1=45$ elements, in contradiction with $|G|=30$.

Therefore $N=1$ or $N^{\prime }=1$. In the first case, a 3-Sylow is normal in $G$, and in the second case, this is a 5-Sylow. By part (b), $G$ is solvable.