## 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.

### Solutions

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$. 1.b. 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. 2. 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$H/H\cap K \cong HK/K$. 3.a. 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)\$).

3.b.

Sorry, not available yet.

3.c.

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4.a.

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4.b.

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4.c.

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5.

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