1. Let $G$ be a finite group and let $p$ be a prime number.
   1. Let $C(g)$ be the centralizer of $g ∈ G$, and let $Z$ be the centre of $G$.
      1. State a formula relating the sizes of $C(g), Z$ and $G$.
      2. Prove that any group of order $p^n$, for $n>0$, has a non-trivial centre.
      3. Prove that if the quotient group $G / Z$ is cyclic then $G$ is abelian.
      4. Prove that every non-abelian group of order $p^3$ has centre of order $p$.
   2. 1. Explain why the group of automorphisms $\operatorname{Aut}(G)$ of $G$ is isomorphic to a subgroup of the symmetric group on $n-1$ letters where $n$ is the order of $G$.
      2. What is $\operatorname{Aut}(G)$ when $G$ is the cyclic group of order $p$ ?
      3. Prove that if $G$ is a simple and $H$ is a proper subgroup of index $n$ then $G$ has at most $n !$ elements.
   3. 1. State the three Sylow Theorems.
      2. Find all groups, up to isomorphism, of order 21.
      3. Find all groups, up to isomorphism, of order 325.

Solution

1. 1. 1. Class equation: ${|G|}={|Z|}+\sum_{g∈Δ}\left|G:C(g)\right|$ where Δ is a set of representatives of conjugacy classes having more than one element.
      2. As $e∈Z,{|Z|}≥1$. By Lagrange's Theorem, for $g∈Δ$, $p\big|{|G:C(g)|}$. As $p\big|{|G|}$, using (i), $p\big|{|Z|}$.
      3. If $G/Z$ is cyclic then $∃g∈G:gZ$ generates $G/Z$. But then $G=\Set{g^kz|k≥0,z∈Z}$ and $(g^kz)(g^{k'}z')=g^{k+k'}zz'=(g^{k'}z')(g^kz)$ as $z,z'∈Z$, so $G$ is abelian.
      4. By Lagrange ${|Z|}=1,p,p^2,p^3$. Since $G$ is non-abelian ${|Z|}≠p^3$. By (ii) ${|Z|}≠1$. By (iii), ${|Z|}≠p^2$ (for then ${|G/Z|}≅C_p$). Hence ${|Z|}=p$.
   2. 1. Every element $ϕ∈\operatorname{Aut}(G)$ satisfies $ϕ(e)=e$ and permutes the remaining $n-1$ elements. Hence $\operatorname{Aut}(G)$ is a subgroup of $\operatorname{Sym}(G∖\{e\})≅\text{Sym}_{n-1}$.
      2. $\operatorname{Aut}(C_p)=C_{p-1}$
      3. $G$ acts by left multiplication on the cosets $G/H$. As ${|G:H|}=n$ this defines a group homomorphism $ϕ:G→\text{Sym}_n$. If ${|G|}>n!={|\text{Sym}_n|}$, then $ϕ$ has non-trivial kernel, let $x∈\kerϕ∖\{e\}$. Then $xgH=gH$ and $g^{-1}xg∈H$. Then the normal subgroup generated by $x$\[⟨⟨x⟩⟩=\left< g^{-1}xg\middle|g∈G\right>\]is contained in $H$. So $⟨⟨x⟩⟩$ is a proper normal subgroup of $G$, contradicting that $G$ is simple.
   3. 1. Sylow Theorem: Let ${|G|}=p^αm$ with $p∤m$.
         • ∃ subgroup $S_p$ of order $p^α$.
         • All such subgroup are conjugate.
         • If $n_p$ is the number of such subgroups, then $n_p≡1\pmod p$ and $n_p|m$.
      2. Groups of Order 21 $n_7≡1\pmod7,n_7|3$ so $S_7$ is unique and normal.
         So $S_3⋅S_7$ is the whole group and $G$ is a semidirect product by $ϕ:S_3→\operatorname{Aut}(S_7)≅C_6$. The two non-trivial ϕ differ by an automorphism of $S_3$, $\operatorname{Aut}(C_3)=C_2$. Hence there are two groups up to isomorphism of order 21: $C_3×C_7≅C_{21}$ and $C_7⋊_ϕC_3,ϕ(a)=a^2$
      3. $325=5^2⋅13$
         $n_{13}≡1\pmod{13}$ and $n_{13}|25⇒n_{13}=1$
         $n_5≡1\pmod5$ and $n_5|13⇒n_5=1$
         Thus $G=S_5×S_{13}$
         But $S_{13}≅C_{13}$ and $S_5≅C_{25}$ or $≅C_5×C_5$.
         So there are two groups up to isomorphism of order 325: $C_{25}×C_{13}≅C_{325}$ or $C_5×C_5×C_{13}≅C_5×C_{65}$