nonabelian-group &laquo; WordPress.com Tag Feed http://en.wordpress.com/tag/nonabelian-group/ Feed of posts on WordPress.com tagged "nonabelian-group" Mon, 20 May 2013 08:46:24 +0000 http://en.wordpress.com/tags/ en <![CDATA[Prove that a group with a given presentation is infinite]]> http://crazyproject.wordpress.com/2010/07/27/prove-that-a-group-with-a-given-presentation-is-infinite/ Tue, 27 Jul 2010 15:00:35 +0000 nbloomf http://crazyproject.wordpress.com/2010/07/27/prove-that-a-group-with-a-given-presentation-is-infinite/ Prove that G = \langle x,y \ |\ x^3 = y^3 = (xy)^3 = 1 \rangle is an infinite group as follows. Let p be a prime congruent to 1 mod 3 and let G_p denote the nonabelian group of order 3p. Let a,b \in G_p with |a| = p and |b| = 3. Prove that ab and ab^2 have order 3. Deduce that G_p is a homomorphic image of G, and from this deduce that G is infinite, using the fact that there are infinitely many primes congruent to 1 mod 3. [Note that every nonidentity element of G_p has order 3 or p.]

Let p be a prime congruent to 1 mod 3, and let G_p = Z_p \rtimes_\varphi Z_3, where Z_p = \langle a \rangle, Z_3 = \langle b \rangle, and \varphi(b)(a) = \gamma(a) for some order 3 automorphism \gamma \in \mathsf{Aut}(Z_p).

In particular, \gamma(a) = a^k for some k \not\equiv 1 mod p, and k^3 \equiv 1 mod p. Note that k^2 + k + 1 = (k^3-1)/(k-1) \equiv 0/(k-1) \equiv 0 mod p.

Now we compute (a,b)^3: (a,b)(a,b)(a,b) = (a \varphi(b)(a), b^2)(a,b) = (a \varphi(b)(a) \varphi(b^2)(a), b^3) = (a \cdot a^k \cdot a^{k^2}, b^3) = (a^{1+k+k^2},b^3) = (1,1). Thus |(a,b)| = 3.

Similarly, it is easy to see that |(a,b^2)| = 3.

Now let S = \{x,y\} and \theta : S \rightarrow G_p be defined by \theta(x) = ab, \theta(y) = b. Clearly ab and b generate G_p. By the universal property of free groups there exists a unique group homomorphism \Psi : F(S) \rightarrow G_p extending \theta; moreover, because (as computed above) ab and b satisfy b^3 = (ab)^3 = (abb)^3 = 1, there exists a unique group homomorphism \Phi : G \rightarrow G_p such that \Phi \circ pi = \Psi. Since \Psi is surjective, so is \Phi. Thus G_p is a homomorphic image of G.

Since there are infinitely many primes congruent to 1 mod 3, G cannot have finite order.

]]> <![CDATA[Free groups of rank at least 2 are nonabelian]]> http://crazyproject.wordpress.com/2010/07/23/free-groups-of-rank-at-least-2-are-nonabelian/ Fri, 23 Jul 2010 15:00:57 +0000 nbloomf http://crazyproject.wordpress.com/2010/07/23/free-groups-of-rank-at-least-2-are-nonabelian/ Prove that if |S| > 1 then F(S) is nonabelian.

First we show that the free group of rank 2 is nonabelian. Suppose to the contrary that F(S) is abelian where |S| = 2; say S = \{a,b\}. Define \varphi : S \rightarrow D_{2n} by \varphi(a) = r, \varphi(b) = s. By the universal property of free groups, there exists a unique group homomorphism \Phi : F(S) \rightarrow D_{2n}. This homomorphism is clearly surjective, so that by the First Isomorphism Theorem, D_{2n} \cong F(S)/\mathsf{ker}\ \Phi. However, every quotient of an abelian group is abelian, but if n \geq 3, D_{2n} is not abelian. Thus F(S) is nonabelian.

Clearly every free group of rank at least 2 contains a subgroup which is free of rank 2. (For instance, use the universal property.) Thus no free group of rank at least 2 can be abelian.

]]> <![CDATA[Some properties of nonabelian p groups of order p³]]> http://crazyproject.wordpress.com/2010/07/02/some-properties-of-nonabelian-p-groups-of-order-p%c2%b3/ Fri, 02 Jul 2010 17:00:28 +0000 nbloomf http://crazyproject.wordpress.com/2010/07/02/some-properties-of-nonabelian-p-groups-of-order-p%c2%b3/ Prove that if p is a prime and P a nonabelian group of order p^3, then |Z(P)| = p and P/Z(P) \cong Z_p \times Z_p.

By Lagrange, there are 4 possibilities for |Z(P)|: 1, p, p^2, and p^3. However, we know that Z(P) is nontrivial, and if |Z(P)| = p^3, then P is abelian.

If |Z(P)| = p^2, then P/Z(P) \cong Z_p is cyclic, and we have P abelian, a contradiction.

If |Z(P)| = p, there are two possibilities for the isomorphism type of P/Z(P): Z_{p^2} and Z_p \times Z_p. In the first case, P is abelian. Thus P/Z(P) \cong Z_p \times Z_p.

]]> <![CDATA[Characterize the normal subgroups of a direct product of nonabelian simple groups]]> http://crazyproject.wordpress.com/2010/06/23/characterize-the-normal-subgroups-of-a-direct-product-of-nonabelian-simple-groups/ Wed, 23 Jun 2010 17:00:25 +0000 nbloomf http://crazyproject.wordpress.com/2010/06/23/characterize-the-normal-subgroups-of-a-direct-product-of-nonabelian-simple-groups/ Let K_1, …, K_n be nonabelian simple groups and let G = K_1 \times \cdots \times K_n. Prove that every normal subgroup of G is of the form G_I for some subset I \subseteq \{1,2,\ldots,n\} (where G_I is as defined in this previous exercise). [Hint: If N \leq G is normal and x = (a_i) \in N with some a_i \neq 1, show that there exists g_i \in G_i not commuting with a_i. Show then that [(1,\ldots,1,g_i,1,\ldots,1),x] \in K_i \cap N’ title=’[(1,\ldots,1,g_i,1,\ldots,1),x] \in K_i \cap N’ class=’latex’ /> and deduce that <img src=.]

Suppose N \leq G is normal and nontrivial, and let x = (a_i) \in N where a_k \neq 1 for some k. Since K_k is simple and nonabelian, there exists g_k \in K_k such that g_ka_k \neq a_kg_k, as otherwise Z(K_k) is nontrivial.

Let y = (b_i), where b_i = g_k if i = k and 1 otherwise.

Consider [y,x] = y^{-1}x^{-1}yx; note that since N is normal, y^{-1}x^{-1}y \in N, hence [y,x] \in N. Moreover, [y,x]_i = g_k^{-1}a_k^{-1}g_ka_k if i=k and 1 otherwise, so that [y,x] \in G_{\{k\}}. Thus [y,x] \in G_{\{k\}} \cap N.

Now G_{\{k\}} \cap N is a nontrivial normal subgroup of G_{\{k\}} \cong K_k, which is simple. Thus G_{\{k\}} \cap N = G_{\{k\}}, and we have G_{\{k\}} \leq N.

Thus N = G_I, where I = \{ i \ |\ a_i \neq 1\ \mathrm{where}\ x_i = a_i\ \mathrm{for\ some}\ x \in N \}.

]]> <![CDATA[In a nonabelian p-group of order p³, the commutator subgroup and center are equal]]> http://crazyproject.wordpress.com/2010/06/20/in-a-nonabelian-p-group-of-order-p%c2%b3-the-commutator-subgroup-and-center-are-equal/ Sun, 20 Jun 2010 18:00:25 +0000 nbloomf http://crazyproject.wordpress.com/2010/06/20/in-a-nonabelian-p-group-of-order-p%c2%b3-the-commutator-subgroup-and-center-are-equal/ Prove that if p is a prime and P a nonabelian group of order p^3, then P^\prime = Z(P).

Since P is nonabelian, we have Z(P) \neq P. Moreover, if |Z(P)| = p^2, then P/Z(P) \cong Z_p is cyclic, so that P is abelian; thus |Z(P)| \neq p^2. Since every p-group has a nontrivial center, we have |Z(P)| = p by Lagrange.

Now |P/Z(P)| = p^2; we know that every group of order p^2 is abelian, thus we have P^\prime \leq Z(P). Since P is nonabelian, P^\prime \neq 1; thus |P^\prime| \geq p, and we have P^\prime = Z(P).

]]> <![CDATA[A nonabelian group whose every proper subgroup is abelian is not simple]]> http://crazyproject.wordpress.com/2010/06/07/a-nonabelian-group-whose-every-proper-subgroup-is-abelian-is-not-simple/ Mon, 07 Jun 2010 17:00:27 +0000 nbloomf http://crazyproject.wordpress.com/2010/06/07/a-nonabelian-group-whose-every-proper-subgroup-is-abelian-is-not-simple/ Prove that if G is any nonabelian group in which every proper subgroup is abelian then G is not simple.

We begin with a lemma.

Lemma: If M \leq G is maximal, then for all g \in G, g^{-1}Mg \leq G is maximal. Proof: Suppose to the contrary that M is maximal and that g^{-1}Mg is not; let g^{-1}Mg \leq H \leq G. Then M \leq gHg^{-1} \leq G, with all containments proper, a contradiction. \square

Let G be a finite nonabelian group in which every proper subgroup is abelian. Suppose that G is simple.

Let M \leq G be a maximal subgroup. By this previous exercise, G \neq \bigcup_{g \in G} g^{-1}Mg; let x \in G such that x is not in a conjugate of M. Now x \in N for some maximal subgroup N \leq G. Note that N is not conjugate to M, so that the conjugates of N and M are pairwise distinct. Moreover, by this previous exercise, the conjugates of N and M intersect trivially pairwise. Thus the nonidentity elements in conjugates of N and M form disjoint subsets of G. By this previous exercise, the number of nonidentity elements in conjugates of M is (|M|-1)[G:M] and in conjugates of N is (|N|-1)[G:N], so that G has at least (|M|-1)[G:M] + (|N|-1)[G:N] nonidentity elements. Now note the following.

|G| > (|M|-1)[G:M] + (|N|-1)[G:N] (strictness since G also contains 1)
 =  (|M|-1)\frac{|G|}{|M|} + (|N|-1)\frac{|G|}{|N|} (since G is finite)
 =  |G| - \frac{|G|}{|M|} + |G| - \frac{|G|}{|N|}
 =  |G|(2 - (\frac{1}{|M|} + \frac{1}{|N|})), so that
|G| > |G|(2 - \frac{|M|+|N|}{|M||N|}) and hence
1 > 2 - \frac{|M|+|N|}{|M||N|}, and thus
|M|+|N| > |M||N|.

Then (without loss of generality) we have |M| \in \{1,2\}, since for natural numbers a and b with a \geq 3, ab \geq a+b.

Suppose |M| = 1. If |G| is composite, then via Cauchy’s theorem we have a contradiction (as there exist proper nontrivial subgroups in G). Then |G| is prime, but then G \cong \mathbb{Z}/(p) is abelian, a contradiction. In particular, we may assume that |M|, |N| \geq 2.

Suppose |M| = 2. Then since M is maximal, M is in fact a Sylow 2-subgroup of G. Hence |G| = 2k where k is odd, and by this previous exercise, G is not simple, a contradiction.

Thus no such group G exists.

]]> <![CDATA[Classify nonabelian groups of order pq, p and q distinct primes]]> http://crazyproject.wordpress.com/2010/05/13/classify-nonabelian-groups-of-order-pq-p-and-q-distinct-primes/ Thu, 13 May 2010 17:00:33 +0000 nbloomf http://crazyproject.wordpress.com/2010/05/13/classify-nonabelian-groups-of-order-pq-p-and-q-distinct-primes/ Let p and q be primes with p < q. Prove that a nonabelian group G of order pq has a nonnormal subgroup of index q, so that there exists an injective homomorphism G \rightarrow S_q. Deduce that G is isomorphic to a subgroup of the normalizer in S_q of the cyclic group generated by (1\ 2\ \ldots\ q).

If Z(G) = G, then G is abelian, a contradiction. If |Z(G)| is p or q, then G/Z(G) is cyclic, so that G is abelian. Hence Z(G) = 1.

Let x \in G be a nonidentity element; note that |x| is either p or q, since if |x| = pq we have G cyclic, a contradiction. Moreover C_G(x) is not all of G, since otherwise x \in Z(G). Since \langle x \rangle \leq C_G(x) \leq G and \langle x \rangle is maximal by Lagrange, the G-conjugacy class containing x has order |G|/|x|. That is, every nonidentity element is in a conjugacy class containing either p elements (if |x| = q) or q elements (If |x| = p). Thus the class equation for G is |G| = 1 + mp + nq, where m is the number of order p conjugacy classes and n the number of order q conjugacy classes.

Now G has a subgroup H of order p by Cauchy’s Theorem. Suppose H is normal; then H is a union of conjugacy classes. However, the smallest possible union of conjugacy classes which contains 1 and some nonidentity element has order p+1. , a contradiction. Thus no subgroup of order p (in particular H) is normal.

Let \varphi : G \rightarrow S_{G/H} denote the permutation representation of G induced by the action of G on the left cosets of H by left multiplication, and let K denote the kernel of this action. Note that K \leq H and that K is normal in G. Since H has prime order, K = 1. Thus \varphi is injective.

Let x \in G be an element of order q by Cauchy’s Theorem. Note that \langle x \rangle is a maximal subgroup of G, and that all conjugates of \langle x \rangle either coincide or intersect trivially, since \langle x \rangle has prime order. If N_G(x) is a proper subgroup of G, we have N_G(\langle x \rangle) = \langle x \rangle. By a previous result, the number of nonidentity elements in G which are in some conjugate of \langle x \rangle is at most pq-p; in fact this number is equal to pq-p since two distinct conjugates intersect trivially. Thus G has at least pq-p elements of order q.

Similarly, since H is not normal, exactly pq-q nonidentity elements of G are in conjugates of H, so that G has at least pq-q elements of order p. Combining these results, we see that |G| \geq 1 + pq-q + pq-p = 1 + 2pq - (p+q) \geq pq+1 > |G|, a contradiction. Thus N_G(x) is in fact all of G.

Suppose now that for some a and b, we have x^aH = x^bH. Then x^{a-b} \in H; because |x| = q and |H| = p are distinct primes, we have x^{a-b} = 1. Thus a = b mod q. Thus the cosets H, xH, x^2H, \ldots, x^{q-1}H are distinct; because G/H is finite, these are precisely the elements of G/H. We can see that \varphi(x) = (H\ xH\ \ldots\ x^{q-1}H). Since \langle x \rangle \vartriangleleft G, we have \langle \varphi(x) \rangle \vartriangleleft \mathsf{im}\ \varphi; hence \mathsf{im}\ \varphi \leq N_{S_{G/H}}(\langle (H\ xH\ \ldots\ x^{q-1}H) \rangle).

]]> <![CDATA[Classify groups of order 6]]> http://crazyproject.wordpress.com/2010/05/02/classify-groups-of-order-6/ Sun, 02 May 2010 18:00:08 +0000 nbloomf http://crazyproject.wordpress.com/2010/05/02/classify-groups-of-order-6/ Prove that every nonabelian group of order 6 has a nonnormal subgroup of order 2. Use this to classify nonabelian groups of order 6.

Let G be a nonabelian group of order 6.

We claim that if x is any element of order 2 and y is any element of order 3, then x and y do not commute. Proof of claim: Suppose otherwise that xy = yx. Then |xy| = 6, and we have G = \langle xy \rangle. Thus G is cyclic, hence abelian, a contradiction.

Now by Cauchy’s Theorem, there exist x,y \in G such that |x| = 2 and |y| = 3, and xy \neq yx. Thus y\langle x \rangle = \{ y, yx \} \neq \{ y, xy \} = \langle x \rangle y, so that \langle x \rangle is not normal.

Now G acts on the left cosets of \langle x \rangle by left multiplication. Let \varphi : G \rightarrow S_3 be the permutation representation induced by this action. Certainly \langle x \rangle = \mathsf{stab}(\langle x \rangle), so that \mathsf{ker}\ \varphi \leq \langle x \rangle. Now \langle x \rangle is a nonnormal subgroup of prime order and \mathsf{ker}\ \varphi is normal, so that \mathsf{ker}\ \varphi = 1. Thus \varphi is injective. Because |G| = |S_3| = 6, \varphi is an isomorphism. Thus G \cong S_3.

]]> <![CDATA[In a nonabelian group, the set of all elements of a given order is not necessarily a subgroup]]> http://crazyproject.wordpress.com/2010/01/23/in-a-nonabelian-group-the-set-of-all-elements-of-a-given-order-is-not-necessarily-a-subgroup/ Sat, 23 Jan 2010 22:42:50 +0000 nbloomf http://crazyproject.wordpress.com/2010/01/23/in-a-nonabelian-group-the-set-of-all-elements-of-a-given-order-is-not-necessarily-a-subgroup/ Let n \geq 3. Show that \{ x \in D_{2n} \ |\ x^2 = 1\} is not a subgroup of D_{2n}.

We know that every element of D_{2n} which is not a power of r has order 2; in particular, sr and sr^2 have order 2. But srsr^2 = ssr^{-1}r^2 = r has order n \geq 3. Thus this set is not closed under multiplication.

]]> <![CDATA[Demonstrate that right multiplication is not a left group action of a group on itself]]> http://crazyproject.wordpress.com/2010/01/21/demonstrate-that-right-multiplication-is-not-a-left-group-action-of-a-group-on-itself/ Thu, 21 Jan 2010 06:48:23 +0000 nbloomf http://crazyproject.wordpress.com/2010/01/21/demonstrate-that-right-multiplication-is-not-a-left-group-action-of-a-group-on-itself/ Let G be a nonabelian group. Show that the action of G on itself by g \cdot a = ag does not satisfy the axioms of a left group action.

Since G is nonabelian, there exist g_1, g_2 \in G such that g_1 g_2 \neq g_2 g_1. Then we have g_1 \cdot (g_2 \cdot 1) = g_1 \cdot 1g_2 = (1g_2)g_1 = 1(g_2g_1) \neq 1(g_1g_2) = (g_1g_2) \cdot 1. Thus this mapping fails to be a group action.

]]> <![CDATA[General linear groups of dimension at least 2 are nonabelian]]> http://crazyproject.wordpress.com/2010/01/16/general-linear-groups-of-dimension-at-least-2-are-nonabelian/ Sat, 16 Jan 2010 14:00:46 +0000 nbloomf http://crazyproject.wordpress.com/2010/01/16/general-linear-groups-of-dimension-at-least-2-are-nonabelian/ Show that GL_n(F) is nonabelian for all n \geq 2 and all fields F.

Recall that every field contains 0 and 1, and that 0 \neq 1. Suppose now that A, B are matrices in GL_n(F) such that the first row of A is [1,0,\ldots,0,1], the first column of A is [1,0,\ldots,0,0], the first row of B is [1,0,\ldots,0,0], and the first column of B is [1,0,\ldots,0,1]. Such matrices always exist in GL_n(F); for instance, take the identity matrix and change the (1,n) or (n,1) entry from 0 to 1. The resulting matrix is either upper or lower triangular, so that the determinant is the product of the diagonal entries. This product is 1, so that A and B are invertible.

With A and B having this form, the (1,1)-entry of AB is 2 and the (1,1)-entry of BA is 1. If 1 = 2 in F then we have 0 = 1, a contradiction. Since matrices are equal precisely when their corresponding entries are equal, we have AB \neq BA. Thus GL_n(F) is nonabelian for n \geq 2 and for all fields F where 0 \neq 1. \blacksquare

]]> <![CDATA[Show that a given general linear group is nonabelian]]> http://crazyproject.wordpress.com/2010/01/15/show-that-a-given-general-linear-group-is-nonabelian/ Fri, 15 Jan 2010 14:00:36 +0000 nbloomf http://crazyproject.wordpress.com/2010/01/15/show-that-a-given-general-linear-group-is-nonabelian/ Show that GL_2(\mathbb{F}_2) is non-abelian.

We have \left[ {1 \atop 1}{1 \atop 0} \right] \cdot \left[ {0 \atop 1}{1 \atop 0} \right] = \left[ {1 \atop 0}{1 \atop 1} \right] and \left[ {0 \atop 1}{1 \atop 0} \right] \cdot \left[ {1 \atop 1}{1 \atop 0} \right] = \left[ {1 \atop 1}{0 \atop 1} \right], so GL_2(\mathbb{F}_2) is non-abelian.

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