- Prove that there exists a unique subgroup $H$ such that $\left| H \right| = q$.
- Prove that $H$ is a normal subgroup of $G$.
- Examine if $G$ is a solvable group.
On Solvable Sylow groups
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On Solvable Sylow groups
Let $p, q$ be prime numbers such that $p<q$ and let $G$ be a group such that $\left| G \right| =pq$.
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- Community Team
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Re: On Solvable Sylow groups
Hello.
i. Firstly, there exist \(\displaystyle{q}\) - Sylow subgroups of \(\displaystyle{G}\).
If \(\displaystyle{n_{q}}\) measures the \(\displaystyle{q}\) - Sylow subgroups of \(\displaystyle{G}\), then,
\(\displaystyle{n_{q}\equiv 1 mod(q)}\) and \(\displaystyle{n_q\mid p}\), so \(\displaystyle{n_q=1}\).
ii. We have that \(\displaystyle{[G:H]=p}\) and \(\displaystyle{p}\) is the smallest prime number
which divides \(\displaystyle{|G|=p\,q}\), so \(\displaystyle{H\triangleleft G}\).
iii. The answer is "YES" . Consider the solvable series
\(\displaystyle{\left\{e\right\}\triangleleft H\triangleleft G}\) and it holds
\(\displaystyle{H/\left\{e\right\}\cong H\cong \mathbb{Z}_{q}\,\,,G/H\cong \mathbb{Z}_{p}}\).
i. Firstly, there exist \(\displaystyle{q}\) - Sylow subgroups of \(\displaystyle{G}\).
If \(\displaystyle{n_{q}}\) measures the \(\displaystyle{q}\) - Sylow subgroups of \(\displaystyle{G}\), then,
\(\displaystyle{n_{q}\equiv 1 mod(q)}\) and \(\displaystyle{n_q\mid p}\), so \(\displaystyle{n_q=1}\).
ii. We have that \(\displaystyle{[G:H]=p}\) and \(\displaystyle{p}\) is the smallest prime number
which divides \(\displaystyle{|G|=p\,q}\), so \(\displaystyle{H\triangleleft G}\).
iii. The answer is "YES" . Consider the solvable series
\(\displaystyle{\left\{e\right\}\triangleleft H\triangleleft G}\) and it holds
\(\displaystyle{H/\left\{e\right\}\cong H\cong \mathbb{Z}_{q}\,\,,G/H\cong \mathbb{Z}_{p}}\).
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