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\newcommand{\LEMMA}{\vspace{0.05in} \noindent \fbox{\large\bf Lemma:}  }
\newcommand{\COR}{\vspace{0.05in} \noindent \fbox{\large\bf Corollary:}  }
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\newcommand{\THM}{\vspace{0.05in} \noindent \fbox{\large\bf Theorem:}  }
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\noindent
{\bf Math 4720 \& 5210}
\hfill {\Large \bf Uniquely Simple} \hfill 
{\bf Abstract Algebra}

\vspace{0.05in}

\THM Let $G$ be a simple group with $|G|>2$ and assume $G$ acts on a finite set $X$.\\ \force \hfill Then $G$ acts as even permutations on $X$.

\vspace{-0.05in}

\bproof   Let $G$ acts on $X$. We then have the corresponding permutation representation $\varphi:G \to S(X)$ where $\varphi(g)[x] = g \bdot x$. Since $G$ is simple $\mathrm{Ker}(\varphi)$ must be $\{1\}$ or $G$. If $\mathrm{Ker}(\varphi)=G$, then $G$ acts trivially: each element acts as the identity permutation $(1)$ (which is even). On the other hand, suppose $\mathrm{Ker}(\varphi)=\{1\}$. Then by the first isomorphism theorem, $G \cong \varphi(G)$ a subgroup of $S(X)$. Recall that subgroups of $S(X)$ are either even or half-even and half-odd. Thus either $\varphi(G)$ is even (done) or $\varphi(G)$ is half-even and half-odd. We show the latter case is impossible. If $\varphi(G)$ is half-even, then $\varphi(G)$ intersected with the subgroup of even permutations in $S(X)$ yields a subgroup of $\varphi(G)$ of index 2. By the index 2 theorem, this is a normal subgroup of $\varphi(G)$. But $|G|>2$ so this is a non-trivial (proper) normal subgroup contradicting the assumption that $G$ is simple. \eproof

\vspace{-0.05in}

\COR Let $G$ be a non-Abelian simple group. Suppose $G$ has a subgroup of (finite) index $k>1$. Alternatively, suppose $G$ acts non-trivially on a set of size $k$. Then $G$ is isomorphic to a subgroup of $A_k$ where $k \geq 5$.

\vspace{-0.05in}

\bproof  Non-Abelian implies $|G|>2$.  Let $H$ be a subgroup of $G$ of index $[G:H]=k$. Then $G$ acts non-trivially (by left multiplication) on the (left) cosets of $H$ in $G$: $g \bdot xH = (gx)H$. Thus $G$ acts non-trivially on a set of size $[G:H]=k$ and so $G$ is isomorphic to a subgroup of $A_k$. Finally, the alternating groups $A_k$ for $k<5$ do not contain non-Abelian simple groups ($A_2$ is trivial, $A_3$ is cyclic order 3, and $A_4$ is not simple and its proper subgroups are Abelian). Therefore, $k \geq 5$. \eproof

It turns out that there are many interesting {\it infinite} simple groups (which are necessarily non-Abelian). The above corollary implies that infinite simple groups cannot have subgroups of finite index.

%\vspace{0.05in}

\EX There are no simple groups of order $80 = 2^4 \cdot 5$. Suppose $G$ is such a group. By the third Sylow theorem, the number of Sylow 2-subgroups, $n_2$, must divide 5. Since $G$ is simple, $n_2 \not=1$ (otherwise, the unique Sylow 2-subgroup would be a proper non-trivial normal subgroup). Therefore, $n_2=5$. Now $G$ acts on the set of Sylow 2-subgroups via conjugation. By the second Sylow theorem, this action has a single orbit. Thus it is a non-trivial action on set of size 5. Therefore, $G$ must be isomorphic to a subgroup of $A_5$. But this is impossible since $|G|=80 > |A_5|=60$. 

%\vspace{0.05in}

\THM There is a unique (up to isomorphism) simple group of order 60, namely $A_5$.

\vspace{-0.05in}

\bproof  Let $G$ be simple and $|G|=60=2^2 \cdot 3 \cdot 5$. Let $\mathcal{S}$ denote the set of Sylow 2-subgroups of $G$ and let $n_2=|\mathcal{S}|$. The third Sylow theorem says that $n_2 \equiv 1$ (mod 2) and $n_2$ divides 15. Therefore, $n_2$ is either 1, 3, 5, or 15. Since $G$ is simple, $n_2 \not=1$ (otherwise the unique Sylow 2-subgroup would be normal contradicting the simplicity of $G$).

Recall that $G$ acts on $\mathcal{S}$ via conjugation (the conjugate of a Sylow 2-subgroup is a Sylow 2-subgroup). Also, the second Sylow theorem says that this action is transitive (i.e., all of $\mathcal{S}$ is a single orbit). Thus since $|\mathcal{S}|=n_2>1$, this is a non-trivial action. So $G$ is a simple group which acts non-trivially on a set of size $n_2$.

If $n_2=3$, the theorems above would imply that $G$ is isomorphic to a subgroup of $A_3$. But this is impossible since $G$ is too big: $|G|=60>3=|A_3|$.
If $n_2=5$, the same theorems would imply that $G$ is isomorphic to  a subgroup of $A_5$, say $G \cong \overline{G} \subseteq A_5$. But $|G|=|\overline{G}|=60$ and $|A_5|=60$ so $G \cong \overline{G}=A_5$ (and we would be done).

We now turn to the only remaining case: $n_2=15$. Let $P \in \mathcal{S}$. Then $[G:N_G(P)]=n_2=15$ where $N_G(P) = \{ g \in G \;|\; gxg^{-1} \in P \text{ for all } x \in P\}$ is the normalizer of $P$ in $G$. Thus $|N_G(P)|=60/15=4$ and since $P \subseteq N_G(P)$ where $|P|=4$ (since $P$ is a Sylow 2-subgroup) we must have $P=N_G(P)$ (i.e., $P$ is ``self-normalizing'').

Let $x \in G$ be an element of order 2 (these exist by Cauchy's theorem). Notice that $xPx^{-1}=P$ if and only if $x \in N_G(P)=P$. By the second Sylow theorem, $\langle x \rangle$ (a subgroup of order 2) must be contained in some Sylow 2-subgroup, say $P$. Consider $x$ acting on $\mathcal{S}$ via conjugation: $x$ fixes $P$. Since $x$'s order is 2, it can either fix or exchange pairs of elements in $\mathcal{S}$. If $x$ did not fix any element of $\mathcal{S}$ other than $P$, it would act as a permutation consisting of a 1-cycle and 7 (disjoint) transpositions (since $|\mathcal{S}|=15$). This would mean that $x$ acted as an odd permutation. But by our theorem this is  impossible since $G$ is simple. Therefore, $x$ must fix at least one other Sylow 2-subgroup, say $Q$ and so $x \in N_G(Q)=Q$. Therefore, $x \in P \cap Q$. Since $P$ and $Q$ are distinct, order 4, and share $1$ and $x$, we have $|P \cap Q|=2$ (i.e., $P \cap Q = \langle x \rangle = \{1,x\}$). 

Now consider $N=N_G(P \cap Q)$. Now $P$ and $Q$ are of order 4, thus they are Abelian. Therefore, $P \cap Q$ is not just a subgroup but a normal subgroup of both $P$ and $Q$. Therefore, we have both $P$ and $Q$ are contained in $N$ (the normalizer of $P\cap Q$). Therefore, $|N| \geq |P \cup Q| \geq 6$ and since $P$ is a subgroup of $N$ and $N$ is a subgroup of $G$, $|N|$ must be a multiple of $4$ and a divisor of $60$. Thus $|N|=12,20,$ or $60$. We cannot have $|N|=60$ since otherwise, $N_G(P \cap Q)=N=G$ which means $P \cap Q$ is a non-trivial (since $|P\cap Q|=2$) proper normal subgroup (contradicting the simplicity of $G$). If $|N|=20$, then $[G:N]=60/20=3$ and so $G$ would necessarily be isomorphic to a subgroup of $A_3$ (this is impossible since $G$ is too big). Therefore, $|N|=12$. Thus $[G:N]=60/12=5$. Therefore, $G$ is isomorphic to a subgroup of $A_5$. Again, since $G$ has order $60=|A_5|$, this subgroup must be all of $A_5$.

Therefore, in all cases, we must conclude that $G \cong A_5$. \eproof


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