Archive for the ‘Groups and Fields’ Category

For any set X, we denote by {\rm{Sym}}(X) the group of bijective maps from X to X.

Problem. Let \alpha : \mathbb{N} \longrightarrow [1, \infty) be any map which satisfies the following two conditions: \alpha(p) \geq p and \alpha(p+q) \geq \alpha(p) \alpha(q) for all p,q \in \mathbb{N}. Let X be a set with |X|=n. Prove that if G is an abelian subgroup of {\rm{Sym}}(X), then |G| \leq \alpha(n).

Solution.  The proof is by induction on n. If n=1, then |G|=1 \leq \alpha(1). Let X be a set with |X|= n \geq 2 and suppose that the claim is true for any set of size < n. Let G be an abelian subgroup of {\rm{Sym}}(X). Clearly gx=g(x), \ g \in G, x \in X, defines an action of G on X. Let X_1, \ldots , X_k be the orbits corresponding to this action and consider two cases.

Case 1. k=1: Fix an element x_1 \in X. Then X=X_1=Gx_1. Suppose that g_1x_1=x_1 for some g_1 \in G and let x \in X. Then x=gx_1 for some g \in G. Thus, since G is abelian, we have


Hence g_1x=x for all x \in X and thus g_1=1. So the stabilizer of x_1 is trivial and therefore, by the orbit-stabilizer theorem, |G|=|X|=n \leq \alpha(n).

Case 2k \geq 2: Let |X_i|=n_i, \ i=1,2, \ldots, k. Clearly \sum_{i=1}^k n_i=n and, since k \geq 2, we have n_i < n for all i. For every g \in G and 1 \leq i \leq k let g_i=g|_{X_i}, the restriction of g to X_i, and put

G_i=\{g_i: \ g \in G\}.

Then g_i \in {\rm{Sym}}(X_i) and G_i is an abelian subgroup of {\rm{Sym}}(X_i). Thus, by the induction hypothesis

|G_i| \leq \alpha(n_i),

for all i. Now, define \varphi : G \longrightarrow \bigoplus_{i=1}^k G_i by \varphi(g)=(g_1, g_2, \ldots , g_k) for all g \in G. It is obvious that \varphi is one-to-one and so

|G| \leq |\bigoplus_{i=1}^k G_i|=\prod_{i=1}^k |G_i| \leq \prod_{i=1}^k \alpha(n_i) \leq \alpha(\sum_{i=1}^k n_i)=\alpha(n). \ \Box

Remark. The map \alpha: \mathbb{N} \longrightarrow [1, \infty) defined by \alpha(p)=3^{p/3}, for all p \in \mathbb{N}, satisfies both conditions in the above Problem. So if |X|=n and if G is an abelian subgroup of {\rm{Sym}}(X), then |G| \leq 3^{n/3}.


Problem. Let F be a field and suppose that f(x),g(x), p(x) are three polynomials in F[x]. Prove that if both f(x) and p(x) are irreducible and p(x) \mid f(g(x)), then \deg f(x) \mid \deg p(x).

Solution. Let \mathfrak{m} and \mathfrak{n} be the ideals of F[x] generated by f(x) and p(x), respectively. Let E=F[x]/\mathfrak{m} and L = F[x]/\mathfrak{n}. Since both f(x) and p(x) are irreducible, E and L are field extensions of F. Now, define the map \varphi : E \longrightarrow L by \varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n}, for all h(x) \in F[x]. We first show that \varphi is well-defined. To see this, suppose that h(x) \in \mathfrak{m}. Then h(x)=f(x)u(x) for some u(x) \in F[x] and hence h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n}, because p(x) \mid f(g(x)). So \varphi is well-defined. Now \varphi is clearly a ring homomorphism and, since E is a field and \ker \varphi is an ideal of E, we must have \ker \varphi = \{0\}. Therefore we may assume that F \subseteq E \subseteq L and hence \deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box

Throughout G is a group with the center Z(G) and the commutator subgroup G'. The goal is to prove that if G/Z(G) is finite, then G' is finite too. We will also find an upper bound for |G'| in terms of |G/Z(G)|.

Notation. For any a,b \in G, we define [a,b]=aba^{-1}b^{-1} and a^b = bab^{-1}.

Problem 1. Let a,b,c \in G.

1) [a,b]=b^ab^{-1}, \ a^a=a, \ ba=a^bb and [a,b]^c=[a^c,b^c].

2) c[a,b]=[a^c,b^c]c.

3) If [G:Z(G)]=n, then [a,b]^{n+1}=[a,b^2][a^b,b]^{n-1}.

Proof. 1) The first three identities are trivial and the fourth one is true because the map f:G \longrightarrow G defined by f(g)=cgc^{-1}=g^c is a group homomorphism.

2) By 1) we have [a^c,b^c]c =[a,b]^cc=c[a,b].

3) So g^n \in Z(G) for all g \in G, because [G:Z(G)]=n. So [a,b]^nb^{-1}=b^{-1}[a,b]^n with 1) give us

[a,b]^{n+1}=[a,b][a,b]^n =b^ab^{-1}[a,b]^n = b^a[a,b]^n b^{-1}=b^a[a,b][a,b]^{n-1}b^{-1}

=b^ab^ab^{-1}[a,b]^{n-1}b^{-1}= (b^2)^ab^{-2}b[a,b]^{n-1}b^{-1}=[a,b^2](b[a,b]b^{-1})^{n-1}=[a,b^2]([a,b]^b)^{n-1}

=[a,b^2][a^b,b^b]^{n-1}=[a,b^2][a^b,b]^{n-1}. \ \Box

Problem 2. Let C=\{[a,b]: \ a,b \in G\}. If [G:Z(G)]=n, then |C| \leq n^2.

Proof. Define the map \varphi : C \longrightarrow G/Z(G) \times G/Z(G) by \varphi([a,b])=(aZ(G),bZ(G)). If we prove that \varphi is one-to-one, we are done because then |C| \leq |G/Z(G) \times G/Z(G)|=n^2. So suppose that \varphi([a,b])=\varphi([c,d]). Then aZ(G)=cZ(G) and bZ(G)=dZ(G) and hence a^{-1}c \in Z(G) and b^{-1}d \in Z(G). Therefore

[a,b]=aba^{-1}b^{-1}=ab(a^{-1}c)c^{-1}b^{-1} =a(a^{-1}c)bc^{-1}b^{-1}=cbc^{-1}b^{-1} =cbc^{-1}(b^{-1}d)d^{-1}

=cb(b^{-1}d)c^{-1}d^{-1}=cdc^{-1}d^{-1}=[c,d]. \ \Box

Schur’s theorem. If [G:Z(G)]=n, then \displaystyle |G'| \leq n^{2n^3}.

Solution. Let C=\{[a,b]: \ a,b \in G\} and c \in G'. Then c = c_1c_2 \ldots c_m, where c_i \in C. We will choose the integer m to be as small as possible, i.e. if c = c_1'c_2' \ldots c_k', with c_i' \in C, then k \geq m. Now, we know from Problem 2, that the number of elements of C is at most n^2. So if we prove that each c_i can occur at most n times in c = c_1c_2 \cdots c_m, then m \leq n|C| \leq n^3 and thus there will be at most (n^2)^m \leq n^{2n^3} possible values for c and the problem is solved. To prove that each c_i can occur at most n times in c = c_1c_2 \ldots c_m, suppose, to the contrary, that, say c_j, occurs r \geq n+1 times in the product. Then by part 2) of Problem 1, we can move each c_j to the right hand side of the product to get c = c_1'c_2' \ldots c_s' c_j^r, where c_i' \in C and r + s = m. But by part 3) of Problem 1, c_j^{n+1} is a product of n elements of C and hence, since c_j^r=c_j^{r-(n+1)}c_j^{n+1}, we see that c_j^r is a product of r-1 elements of C. Therefore c is a product of s+r-1=m-1 elements of C, which contradicts the minimality of m. \ \Box

If G is finitely generated, then the converse of Schur’s theorem is also true, i.e. if G' is finite, then G/Z(G) is finite too. It’s not hard, try to prove it!

Let G be a group, N a normal subgroup of G and P=\{g_iN: \ i \in I \} the set of cosets of N. Then P is a partition of G and g_iN g_jN = g_ig_j N \in P for all i,j \in I. We’d like to consider the converse of this.

Problem. Let G be a group and suppose that P is a set partition of G which satisfies the following condition:

(*) for every Q_1,Q_2 \in P, there exists Q \in P such that Q_1Q_2 \subseteq Q.

Let N be the element of P which contains the identity element of G. Prove that N is a normal subgroup of G and P is the set of cosets of N in G.

Solution. Notice that an obvious result of (*) is that if Q_1 \in P and g, h \in G, then gQh \subseteq Q_2 for some Q_2 \in P. Now, if a \in N, then a = a \cdot 1 \in N^2 and so N \subseteq N^2. We also have N^2 \subseteq Q for some Q \in P, by (*). Thus N \subseteq Q and so N=Q. Therefore N=N ^2, i.e. N is multiplicatively closed. Let a \in N. Then Na^{-1} \subseteq Q for some Q \in P and since 1 \in Na^{-1}, we get 1 \in N \cap Q. Thus Q=N and so Na^{-1} \subseteq N. Hence a^{-1} \in N and so N is a subgroup. To prove that N is normal, let g \in G. Then, by (*), there exists Q \in P such that gNg^{-1} \subseteq Q. But then 1 \in Q \cap N, because 1 \in gNg^{-1} \subseteq Q, and so Q=N. Thus gNg^{-1} \subseteq N, i.e.  N is normal. Finally, let g \in G and choose Q, Q_1 \in P such that g \in Q and gN \subseteq Q_1. Then g \in Q_1 \cap Q, because g \in Ng \subseteq Q_1, and so Q_1=Q. Hence g N \subseteq Q. Let Q_2 \in P be such that N \subseteq g^{-1}Q \subseteq Q_2. Then 1 \in Q_2 \cap N, because 1 \in g^{-1}Q \subseteq Q_2, and so Q_2=N. Hence gN=Q. \ \Box

As usual, for a subgroup H of a group G, we will denote by N(H) and C(H) the normalizer and the centralizer of H in G. If H = \langle a \rangle, then we will write N(a) and C(a) for N(H) and C(H) respectively.

Problem. Let p be a prime number and let G be a group of order p^n(p+1), \ n \geq 1. Prove that G is not simple.

Solution. If G is abelian, there is nothing to prove. So we suppose that G is non-abelian and simple and we will get a contradiction. By Sylow theorem, the number of Sylow p-subgroups is p+1 and so P=N(P) for every Sylow p-subgroup P of G. Now, we consider two cases.

Case 1 . n=1. Let P be a Sylow p-subgroup. Then |P|=p and so P \subseteq C(P). Suppose that P \neq C(P) and choose a \in C(P) \setminus P. Then P \subseteq C(a). Let Q=aPa^{-1}. We have P \cap Q = \{1\} because |P|=|Q|=p and a \notin P=N(P). Thus PQ \subseteq C(a) and so |C(a)| \geq |PQ|=p^2 which implies that C(a)=G because |C(a)| \mid |G|=p(p+1). Thus \langle a \rangle is in the center of G and hence it’s a normal subgroup of G, contradicting our assumption that G is simple. Therefore N(P)=P=C(P) and we are now done by the Burnside’s normal complement theorem.

Case 2.   n \geq 2. The idea for this case is similar to the one we used for case 2 in this problem. Let P and Q be two distinct Sylow p-subgroups of G and put H=P\cap Q. Then

\displaystyle \frac{p^{2n}}{|H|} = |PQ| \leq |G|=p^n(p+1)

which gives us |H|=p^{n-1} because |H| \mid |G|=p^n(p+1). As a result, H is a non-trivial normal subgroup of both P and Q. Therefore PQ \subseteq N(H) and so |N(H)| \geq |PQ|=p^{n+1}. But |N(H)| \mid |G|=p^n(p+1) and so N(H)=G, i.e. H is a normal subgroup of G. This contradicts our assumption that G is simple. \Box

Suppose that p(x) is an irreducible polynomial over some field F. Let E/F be a finite field extension. The following problem investigates the irreducibility of p(x) over E.

Problem. Suppose that E/F is a finite field extension and p(x) \in F[x]. Prove that if p(x) is irreducible over F and \gcd(\deg p(x), [E:F])=1, then p(x) is irreducible over E.

Solution. We may assume, without loss of generality, that p(x) is monic. Let a be a root of p(x) in some extension of E and let q(x) be the minimal polynomial of a over E. Clearly

\deg q(x) \leq \deg p(x). \ \ \ \ \ \ \ (1)

We also have

[E(a):E)][E:F]=[E(a):F(a)][F(a):F]. \ \ \ \ \ \ (2)

It now follows from (2) and \gcd(\deg p(x), [E:F]) = 1 that \deg p(x) \mid \deg q(x) and so by (1)

\deg p(x)=\deg q(x). \ \ \ \ \ \ \ (3)

Let r(x)=q(x)-p(x) \in E[x]. Then \deg r(x) < \deg q(x) by (3) and the fact that p(x) and q(x) are monic. We also have r(a)=0 and thus r(x)=0 by the minimality of q(x). Hence q(x)=p(x). \ \Box

Note that it is possible for p(x) to be irreducible over E but \gcd(\deg p(x), [E:F]) \neq 1. For example consider p(x)=x^2+1, \ F = \mathbb{Q} and E = \mathbb{Q}(\sqrt{2}).

The symmetric group S_3 is an example of a finite non-abelian group in which every proper subgroup is abelian. This group is not simple because its Sylow 3-subgroup is normal. In this post, we’ll show that this is the case for any finite (non-abelian) group all of whose proper subgroups are abelian.

Notation. Let G be a group and let H be a subgroup of G. We will denote by N(H) the normalizer of H in G. We will also define \mathcal{C}(H) to be the union of conjugates of H, i.e. \mathcal{C}(H)=\bigcup_{g \in G} gHg^{-1}.

Lemma 1. Let G be a finite group and let H be a subgroup of G with this property that gHg^{-1} \cap H = \{1\}, for all g \notin H. Then
1) the intersection of two distinct conjugates of H is the identity element;
2) |\mathcal{C}(H)|=|G|-[G:H] + 1 and thus |G \setminus \mathcal{C}(H)|=[G:H]-1.

Proof. There is nothing to prove if H=\{1\}. So suppose that H \neq \{1\}. Clearly H=N(H) because if g \in N(H) \setminus H, then H=gHg^{-1} \cap H = \{1\}, which is a contradiction. Thus the number of conjugates of H is [G:N(H)]=[G:H].
1) Suppose that g_iHg_i^{-1}, \ i=1,2, are two distinct conjugates of H and g \in g_1Hg_1^{-1} \cap g_2Hg_2^{-1}. Then g=g_1h_1g_1^{-1}=g_2h_2g_2^{-1}, for some h_1,h_2 \in H and hence h_2 \in g_2^{-1}g_1 H g_1^{-1}g_2 \cap H. So, by hypothesis, either h_2=1 or g_2^{-1}g_1 \in H. If g_2^{-1}g_1 \in H, then g_1Hg_1^{-1}=g_2Hg_2^{-1}, which is a contradiction. Thus h_2=1 and so g=1.
2) Since every conjugate of H has |H| elements and, by 1), two distinct conjugates of H have only one element in common , we have

|\mathcal{C}(H)|=[G:H](|H|-1)+1=|G|-[G:H]+1. \ \Box

Lemma 2. Let G be a group and let H_1 and H_2 be two abelian subgroups of G. Let H be the subgroup generated by H_1 and H_2. Then H_1 \cap H_2 is a normal subgroup of H.

Proof. Since H_1 and H_2 are abelian, H_1 \cap H_2 is a normal subgroup of both H_1 and H_2. Thus H_1 \cap H_2 is also a normal subgroup of H because an element of H is a finite product of elements of H_1 and H_2. \ \Box

Problem. Let G be a finite non-abelian group in which every proper subgroup is abelian. Prove that G is not simple.

Solution. Suppose, to the contrary, that G is simple and let H be a maximal subgroup of G. Clearly N(H)=H because H \subseteq N(H) and H is a maximal subgroup of G. Let g \notin H. Then g \notin N(H) and so gHg^{-1} \neq H. Hence the subgroup generated by H and gHg^{-1} is G because it contains H strictly. Therefore gHg^{-1} \cap H is a normal subgroup of G, by Lemma 2. Hence gHg^{-1} \cap H= \{1\} and so, by Lemma 1

|G \setminus \mathcal{C}(H)|=[G:H]-1. \ \ \ \ \ \ \ \ (*)

In particular \mathcal{C}(H) \neq G and so we can choose a \notin \mathcal{C}(H). Let K be a maximal subgroup of G which contains a. Then again, exactly as we proved for H, we have gKg^{-1} \cap K = \{1\} and so , by Lemma 1

|\mathcal{C}(K)|=|G|-[G:K]+1. \ \ \ \ \ \ \ \ \ (**)

Now let g_1,g_2 \in G. Note that g_2Hg_2^{-1} is a maximal subgroup of G because H is a maximal subgroup of G. Since a \in K and a \notin \mathcal{C}(H), we have g_1Kg_1^{-1} \neq g_2Hg_2^{-1} and thus the subgroup generated by g_1Kg_1^{-1} and g_2Hg_2^{-1} is G. Therefore g_1Kg_1^{-1} \cap g_2Hg_2^{-1} is a normal subgroup of G, by Lemma 2. Hence g_1Kg_1^{-1} \cap g_2Hg_2^{-1}=\{1\} and so \mathcal{C}(H) \cap \mathcal{C}(K)=\{1\}. Thus we have proved that

\mathcal{C}(K) \setminus \{1\} \subseteq G \setminus \mathcal{C}(H).

It now follows from (*) and (**) and the above inclusion that

|G| \leq [G:H]+[G:K]-1 \leq |G|/2 + |G|/2 - 1 =|G|-1,

which is non-sense. This contradiction shows that G is not simple. \Box

In fact, the result still holds for finite non-cyclic groups because a simple abelian group is cyclic (of prime order).