 Let $G$ be a finite group and let $p$ be a prime number.
 Let $C(g)$ be the centralizer of $g ∈ G$, and let $Z$ be the centre of $G$.
 State a formula relating the sizes of $C(g), Z$ and $G$.
 Prove that any group of order $p^n$, for $n>0$, has a nontrivial centre.
 Prove that if the quotient group $G / Z$ is cyclic then $G$ is abelian.
 Prove that every nonabelian group of order $p^3$ has centre of order $p$.

 Explain why the group of automorphisms $\operatorname{Aut}(G)$ of $G$ is isomorphic to a subgroup of the symmetric group on $n1$ letters where $n$ is the order of $G$.
 What is $\operatorname{Aut}(G)$ when $G$ is the cyclic group of order $p$ ?
 Prove that if $G$ is a simple and $H$ is a proper subgroup of index $n$ then $G$ has at most $n !$ elements.
 State the three Sylow Theorems.
 Find all groups, up to isomorphism, of order 21.
 Find all groups, up to isomorphism, of order 325.
 Class equation: ${G}={Z}+\sum_{g∈Δ}\leftG:C(g)\right$ where Δ is a set of representatives of conjugacy classes having more than one element.
 As $e∈Z,{Z}≥1$. By Lagrange's Theorem, for $g∈Δ$, $p\big{G:C(g)}$. As $p\big{G}$, using (i), $p\big{Z}$.
 If $G/Z$ is cyclic then $∃g∈G:gZ$ generates $G/Z$. But then $G=\Set{g^kzk≥0,z∈Z}$ and $(g^kz)(g^{k'}z')=g^{k+k'}zz'=(g^{k'}z')(g^kz)$ as $z,z'∈Z$, so $G$ is abelian.
 By Lagrange ${Z}=1,p,p^2,p^3$. Since $G$ is nonabelian ${Z}≠p^3$. By (ii) ${Z}≠1$. By (iii), ${Z}≠p^2$ (for then ${G/Z}≅C_p$). Hence ${Z}=p$.

 Every element $ϕ∈\operatorname{Aut}(G)$ satisfies $ϕ(e)=e$ and permutes the remaining $n1$ elements. Hence $\operatorname{Aut}(G)$ is a subgroup of $\operatorname{Sym}(G∖\{e\})≅\text{Sym}_{n1}$.
 $\operatorname{Aut}(C_p)=C_{p1}$
 $G$ 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 nontrivial kernel, let $x∈\kerϕ∖\{e\}$. Then $xgH=gH$ and $g^{1}xg∈H$. Then the normal subgroup generated by $x$\[⟨⟨x⟩⟩=\left⟨g^{1}xg\middleg∈G\right⟩\]is contained in $H$. So $⟨⟨x⟩⟩$ is a proper normal subgroup of $G$, contradicting that $G$ is simple.
 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 $n_pm$.
 Groups of Order 21 $n_7≡1\pmod7,n_73$ 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 nontrivial 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 $C_7⋊_ϕC_3,ϕ(a)=a^2$
 $325=5^2⋅13$
$n_{13}≡1\pmod{13}$ and $n_{13}25⇒n_{13}=1$
$n_5≡1\pmod5$ and $n_513⇒n_5=1$
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 $C_5×C_5×C_{13}≅C_5×C_{65}$