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2000 November 10

  1. a. State the Sylow theorems (existence, conjugacy and number). b. Deduce that every group $G$ of order 140 contains a normal subgroup of index 4. Is $G'$ necessarily abelian?

  2. Let $G$ be a group and $H$ a subgroup of $G$, $K$ a normal subgroup of $G$. Prove that $HK$ is a subgroup of $G$, that $H\cap K$ is a normal subgroup of $H$, and $HK/K \cong H/H\cap K$.

  3. a. State the "class equation" for a finite group $G$ and prove its validity. b. Prove that if a prime $p$ divides the order of $G$, then $G$ contains an element of order $p$. c. Let $|C(x)|$ denote the order of the conjugacy class of the centralizer of $x \in G$, and $k(G)$ denote the number of conjugacy classes of $G$. Prove that $\sum_{x \in G}|C(x)| = k(G) \cdot |G|$.

  4. a. Define "solvable" group. b. Prove (only) one direction of the following statement: If $N$ is a normal subgroup of $G$, then $G$ is solvable if and only if $N$ and $G/N$ are solvable. c. Prove that every group of order 200 is solvable.

  5. List (do not just count) a complete set of representatives of the isomorphism classes of abelian groups of order $p^5$, where $p$ is any prime.


1. a.

Recall, for prime $p$, a group every element of which has order some power ($\geq 0$) of $p$ is called a $p$-group. If $H$ is a $p$-group and $H < G$, then $H$ is called a $p$-subgroup. A maximal $p$-subgroup of $G$---that is, a $p$-subgroup of $G$ contained in no larger $p$-subgroup---is called a Sylow $p$-subgroup of $G$

First Sylow theorem. Let $G$ be a group of order $p^n m$, with $p$ prime, $n\geq 1$ and $(p,m) = 1$. Then for each $1\leq k < n$, $G$ contains a subgroup of order $p^k$ and every subgroup of $G$ of order $p^k$ ($k < n$) is normal in some subgroup of order $p^{k+1}$.

Second sylow theorem. If $H$ is a $p$-subgroup of a finite group $G$, and if $P$ is any Sylow $p$-subgroup of $G$, then there exists $x\in G$ such that $H\leq xPx^{-1}. In particular, any two Sylow $p$-subgroups of $G$ are conjugate.

Third Sylow theorem. If $G$ is a finite group and $p$ a prime, then the number of Sylow $p$-subgroups of $G$ divides $|G|$ and is of the form $kp +1$ for some $k \geq 0$.


The number of Sylow $7$-subgroups must be congruent to $1$ modulo $7$ and divide $140$, hence must be $1$. Call this subgroup $H$. $H$ is then normal in $G$, by the Second Sylow Theorem. Similary, the number of Sylow $5$-subgroups must be congruent to $1$ modulo $5$ and divide $140$, hence must be $1$. Call this (normal) subgroup $K$. The nonidentity elements of $H$ all have order $7$ and the nonidentity elements of $K$ all have order $5$. As $H$ and $K$ are normal in $G$, $HK$ is a subgroup and as $H\cap K = (e)$, $|HK| = 35$. Thus, by Lagrange's Theorem, $HK$ is a subgroup of index $|G:HK| = |G|/|HK| = 4$. If $g \in G$, then $gHKg^{-1}= gHg^{-1}gKg^{-1}\subseteq HK$, since $H$ and $K$ are both normal in $G$. Thus, $HK$ is also normal in $G$.

Every commutator $[a, b]$, where either $a$ or $b$ is in $HK$, is equal to the identity. As $|HK| = 35$, the commutator subgroup $G'$ is of order $1$, $2$, or $4$. Therefore the commutator subgroup is abelian as all groups of such orders are abelian.


Let $hk \in HK$. We show $(hk)^{-1} = k^{-1}h^{-1} \in HK$. Since $K$ is normal, $hk^{-1}h^{-1} =k'\in K$. Therefore, $k^{-1}h^{-1}= h^{-1}k' \in HK$.

Let $hk \in HK$, and $h_1k_1 \in HK$. We show $hkh_1k_1\in HK$. Since $K$ is normal in $G$, there exist $k_2$ and $k_3$ such that $hkh_1k_1= hkh_1k_1h_1^{-1}h_1= h kk_2h_1 = h h_1h_1^{-1}kk_2h_1 = hh_1k_3$, which is clearly in $HK$. Therefore, $HK$ is a subgroup of $G$.

$K$ is normal in $G$ so $N_G(K) = G$. $H \subseteq N_G(H)$ by definition. So $H\cap K \in N_G(H),$ i.e., $H\cap K$ is normal in $H$. Since $K$ is normal in $G$, the quotient group $HK/K$ is well-defined. Define a map $\phi \colon H \to HK/K$ by $\phi(h) = hK$. Then $\phi$ is a homomorphism since for all $h_1, h_2\in H$, we have $\phi(h_1h_2) = h_1K h_2K = h_1 h_2 K$ because as $K\triangleleft G$. The kernel of $\phi$ consists of all elements $h\in H$ such that $hK =K$; that is, $\ker h = H\cap K$. For any element $hk \in HK$, $\phi(hk) = hK$ so $\phi$ is surjective. Therefore, by the first isomorphism theorem for groups, and $H/H\cap K \cong HK/K$.


Theorem. (Class Equation) Suppose $G$ is a group. Then $$|G| = |Z| + \sum_{g\in T} |G : C(g)|,$$ where $Z$ denotes the center of $G$, $C(g):= \{x \in G: xg = gx\}$ is the centralizer of $g$, and $T$ contains one element from each non trivial conjugacy class of $G$.

Corollary. Let $p$ be a prime number. Any finite group of prime power order has a nontrivial center (i.e., $Z\neq (e)$).


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