I am practicing the concepts of solvable groups. I need help making sure my proof is correct and understanding an example from my lecture notes.

Show that every group of order $500$ is solvable. Notation: $\operatorname{Syl}_p(G) = $ set of Sylow $p$-subgroups of $G.$

Since $500 = 5^3 2^2,$ we have that $|\operatorname{Syl}_5(G)| \equiv 1 \mod 5$ and $|\operatorname{Syl}_5(G)| \mid 2^2$. This implies $|\operatorname{Syl}_5(G)| = 1.$ Suppose $P \in \operatorname{Syl}_5(G).$ Then $\operatorname{Syl}_5(G) = \{P\}$ and $P \lhd G.$ If $P$ is a subgroup of order $b < 3,$ then Sylow's first theorem implies that there exists a subgroup $P'$ with order $p^{b+1}.$ Since $\operatorname{Syl}_5(G)$ contains only one element, it must be the case that $|P| = 5^3 = 125.$ Clearly $P$ is a $p$-group, and Sylow's first theorem guarantees a subnormal series $\{1\} \lhd P_1 \lhd \cdots \lhd P$ with factors $P_j/P_{j-1}$ of order $p.$ This implies the factors of $P$ are cylic (and therefore abelian), so $P$ is solvable. The corresponding factor group is $G/P$ and has order $[G:P] = 4 = 2^2,$ hence $G/P$ is abelian group of order $4$ isomorphic to $\mathbb{Z}_4$ and is therefore solvable. Since $P$ and $G/P$ are both solvable, it follows that $G$ is solvable.

Is this correct?

An example from my lecture notes uses the following proposition:

Let $G$ be a finite group and let $H \leqslant G$ with $[G:H] = n$. If $[G:H] \leqslant 4$ and if $\operatorname{core}_G(H) := \bigcap_{a \in G} aHa^{-1} $ is solvable, then $G$ is solvable.

Every group of order $48$ is solvable. Since $48 = 2^4· 3$, there exists a subgroup $S \leqslant G$ of order $16$ by Sylow’s First Theorem. The group $\operatorname{core}_G(S) \leqslant S.$ Therefore it is a $2$-group and is solvable. Moreover, $[G:S] = 3 \leqslant 4.$ Hence $G$ is solvable.

Question: What is a $2$-group? ($p$-group with $p = 2$?) and why does being a $2$-group imply solvability?

  • 1
    $\begingroup$ A $\;p$- group, $\;p\;$ a prime, is one in which all the elements have order a power of $\;p\;$ . if it is finite it is always nilpotent and thus solvable. And yes: your proof looks correct though perhaps a little lengthy $\endgroup$ – Timbuc Mar 17 '15 at 23:58
  • $\begingroup$ oh okay, thanks. $\endgroup$ – St Vincent Mar 18 '15 at 0:00

Your proof looks fine but you could shorten it quite a bit: as before, there's only one single Sylow $\;5$- subgroup $\;P\;$ which is then normal, but then


and since $\;P\;$ is solvable ($\;p$- subgroup, with $\;p=5$) and also $\;G/P\;$ is solvable (in fact, even abelian,), $\;G\;$ is solvable.

  • $\begingroup$ and $G/P$ is solvable since it's a $p$-group? $\endgroup$ – St Vincent Mar 18 '15 at 0:09
  • $\begingroup$ @StVincent Even better: because it is abelian itself! $\endgroup$ – Timbuc Mar 18 '15 at 0:10

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