# Tag Info

0

Let $G$ be the trivial group, for the only finite example.

0

While you can do this easily by checking the conditions for a subgroup, there is a somewhat easier approach using some general facts. The set of permutations of$~G$ (that is bijections $G\to G$) is a group under function composition (this is true for any set), call it $P(G)$. Define a map $\def\ad{\operatorname{ad}}\ad:G\to P(G)$ by $\ad(g)=(h\in G\mapsto ... 3 Update I forget about the inverses, so please check Walter's answer (I could write it also, but I think there is no need). For your first question.$Z(G) = \{a\in G: \forall b \in Gab = ba\}$. So, take a look at$e$. You have$eb = b = be, \forall b \in G \implies e \in Z(G)$. For the second question. We need to use the fact, that the group operation is ... 2 The line$(ab)x = a(bx) = (ax)b = (xa)b = x(ab), ∀x$in$G$implies that$ab$commutes with arbitrary members of$G$and so$ab \in Z(G)$. [We already know$ab \in G$because$G$is a group; you may have misread this part of the proof]. To prove$Z(G)$is a subgroup, you need to show it is closed under the operation and that it is closed under taking of ... 5 Take$G = H \times H \times H \times \cdots$for$H$any nontrivial group. 2 Let$G = \mathbb Z ^ \mathbb N$(with pointwise addition as the product). Then let$f:G \times G \longrightarrow G$be $$f(g,h)(n) = \begin{cases} g(k), &n = 2k \\ h(k), &n = 2k+1 \end{cases}$$ You can verify$f$is an isomorphism. 1 The identity element in$\Bbb R^*$is$1$and not$0$so the kernel is $$\{x\in G: \phi(x)=1\} = \{z \in \Bbb C \setminus \{0\} \ \mid \ |z| = 1 \}$$ This set can either be described as all elements on the unit circle on the complex plane or as the set of all$\{\cos \theta + i \sin \theta \ \mid \ \theta \in \Bbb R\}$. As for the image of$\phi$... 1$H, N < G$subgroups and$H \unlhd G$normal subgroup, then$HN=\{hn : h \in H, n \in n \}$is a subgroup of$G$. Since if$hn , h_1n_1 \in HN$then$hn ( h_1n_1)^{-1}=hn n_1^{-1} h_1^{-1} = h ( n n_1^{-1} h _1^{-1} (n n_1^{-1})^{-1}) (n n_1^{-1})$and since$ H \unlhd G$,$( n n_1^{-1} h _1^{-1} (n n_1^{-1})^{-1}) \in H$, hence$hn ( h_1n_1)^{-1} \in ...

0

Let $G=S_3$, $A=\langle(1,2)\rangle$, $B=\langle(2,3)\rangle$. Then $$AB=\{e,(1,2),(2,3),(1,3,2)\}$$ Note that the inverse of $(1,3,2)$, that is, $(1,2,3)$ is not in $AB$. So, $AB$ is not in general a subgroup. It is, nevertheless, if $G$ is abelian.

1

Note that $e\in AB$. Let $x\in AB$ then $x=ab$ for some $a\in A$ and $b\in B$. Then $x^{-1}=b^{-1}a^{-1}\in BA$ which is generally not equal to $AB$. So inverse of every element in $AB$ may not exist in $AB$ for which $AB$ is not a subgroup.

1

Generally, no. Consider the subgroups of $S_4$ given by: $A = \{e, (1\ 4)\}, B = \{e, (1\ 2 \ 3), (1\ 3\ 2)\}$. Then $AB = \{e, (1\ 4), (1\ 2\ 3), (1\ 3\ 2), (1\ 2\ 3\ 4), (1\ 3\ 2\ 4)\}$ which isn't closed under multiplication: $(1\ 2\ 3\ 4)(1\ 2\ 3\ 4) = (1\ 3)(2\ 4) \not\in AB$.

0

You're right about the fact that all groups are sets of functions, but it is necessary to be more specific. It is in fact true that any group $G$ is equivalent to a subgroup of the group of permutations of the elements of $G$. Let us then represent each element $a \in G$ as a function--specifically, a permutation of the set $G$ itself, which we will denote ...

2

Le be $a,b\in G$. If $ab\cdot x=x\cdot ab=e$, then $x$ is the inverse of $ab$, denoted by casually $(ab)^{-1},$ and this inverse is unique. So, look that $$(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aea^{-1}=aa^{-1}=e.$$ Similarly, $$(b^{-1}a^{-1})(ab)=e.$$ As the inverse is unique, then $(ab)^{-1}=b^{-1}a^{-1}.$

6

The argument essentially goes: If $x$ satisfies $$(ab)\cdot x=e$$ then $x$ is the inverse of $ab$. That is essentially the definition of an inverse - and, by the properties of a group, defines a unique $x$. The proof says: Suppose $x=b^{-1}a^{-1}$. Then $$(ab)\cdot x = abb^{-1}a^{-1}=aa^{-1}=e$$ hence $x=b^{-1}a^{-1}$ is the inverse of $(ab)$. That ...

1

Hint: You can reduce it (semidirect products) to the following: If $\phi$ is an automorphism of a group $G$ that takes a subgroup $H$ to itself: $\phi(H)\subset H$, does that imply $\phi(H) = H\,$?

5

No. If $H = \mathbb{Z}$, it might be the case that $kHk^{-1} = 2 \mathbb{Z} \subsetneq \mathbb{Z}$. In fact this can actually occur in $G = \text{GL}_2(\mathbb{Q})$, where we can take $$H = \left[ \begin{array}{cc} 1 & \mathbb{Z} \\ 0 & 1 \end{array} \right]$$ $$k = \left[ \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right].$$ As Berci ...

2

If $H$ is finite, this is true, simply because of cardinality arguments. In the infinite case, this is false. You can try to come up with a counterexample.

1

Let $G=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$, where $p_1<p_2<\cdots<p_n$ are distinct primes. Let $N_p$ denote the number of Sylow $p$-subgroups and let $N$ denote the total number of Sylow subgroups, i.e. the sum of all $N_{p_i}$. By the fact that $N_p\equiv 1\pmod{p}$ and $N_p||G|$ we must have that $$N_{p_i}\leq \frac{|G|}{p_i^{a_i}}$$ Thus $$N\leq ... 2 Given two distinct subgroups of order 5 H and K, H \cap K = \{e\} (because otherwise the element in common would generate them both). So if all elements except 1 have order 5, the number of elements in the group must be 4n + 1, which 15 isn't. So that means that there must be an element of order 3. Now assume that all elements have order ... 0 For the second part (no Frobenius Reciprocity needed!): let \chi \in Irr(G), H \leq G. Then$$[\chi_H, \chi_H]= \frac{1}{|H|}\sum_{h \in H}|\chi(h)|^2 \leq [G:H]\cdot \frac{1}{|G|} \sum_{g \in G}|\chi(g)|^2=[G:H]\cdot[\chi,\chi]=[G:H].$$Now, if H is abelian, then \chi_H=\sum_{\lambda \in Irr(H)} a_\lambda \lambda, with a_\lambda non-negative ... 2 They are not isomorphic. In (\mathbb Q\setminus \{0\},\cdot) you have an element that is its own inverse (-1). This does not happen in (\mathbb Z,+) 2 No. (\Bbb Z, +) is generated by \{1,-1\} and (\Bbb Q^\times, \cdot) is not finitely generated. 1 We assume the group is not cyclic. If it generates G we are done. Otherwise by Lagrange it has order 3 or 5. If the order is 5 then then let C be the group generated by g. Since the index of C is 3 and 3 is the smallest prime dividing |G| we conclude that C is normal and so we take G\rightarrow \frac{G}{C} the natural isomorphism ... 2 First note that * is well-defined. If (a,b), (x,y) \in G, then a,x \in \Bbb R^*, so ax \in \Bbb R^*, which means that (ax, a^2 y + b) \in \Bbb R^* \times \Bbb R = G. We show that (1, 0) \in G is the neutral element of G:$$\begin{align*} (1,0) * (x,y) &= (1\cdot x, 1^2 y + 0) = (x,y) \\ (x,y) * (1,0) &= (x\cdot 1, x^2 \cdot 0 + y) = ...

4

Proof of associativity: Assume $(a,b), \ (u,v) , \ (x,y)\in G$. Then $$\big((a,b) * (x,y) \big) * (u,v) = (ax,a^2y+b ) * (u,v) = (axu, a^2 x^2v + a^2y+b)$$ On the other hand, $$(a,b) * \big((x,y) * (u,v) \big) = (a,b) * (xu,x^2v + y ) =\big (axu, a^2(x^2 v +y) + b\big) = (axu, a^2x^2 v +a^2y + b ).$$ Clearly two expressions coincide, therefore ...

6

Sure, take $H = \mathbb{Z}/n\mathbb{Z} \times G$. Even though this might seem artificial, this is the most general solution. For example, if $G= \mathbb{Z}/p\mathbb{Z}$, and $n=q$ with $p,q$ prime numbers such that $q\nmid (p-1)$ and $p\nmid (q-1)$, then whatever $H$ is, we have $|H| =pq$. Standard techniques will show that $H = ... 0 Associativity shouldn't pose much of a problem.$((a,b) * (c,d)) * (e,f)= (ac,a^2d + b) * (e,f)= (ace,a^2c^2f + a^2d + b)$Now perform a similar manipulation with$(a,b) * ((c,d) * (e,f))$and you should get the same expression. 2 Seems to me like there is a detail that needs to be explained in your proof. I get that if$a = a$then$a\cdot e = a\cdot e'$. However the next step would be $$a^{-1}\cdot a\cdot e = a^{-1}\cdot a\cdot e'$$ Now$a^{-1}\cdot a$is equal to the identity. Which one though? 1 I only know a proof without Lagrange (or Lagrange in disguise) for an abelian group. Suppose that$G=\{ a_1,\ldots ,a_n\}$, and set$g:=a_1a_2\cdots a_n$. Then for every$x\in G$the map$a_i\mapsto xa_i$is a permutation of$G$, so that $$g=(xa_1)(xa_2)\cdots (xa_n)=x^na_1a_2\cdots a_n=x^ng.$$ This implies$x^{\mid G\mid}=x^n=e$. 0 Let$d=ord(a)$. Then$a^d=e$and$H=\{ e,a,a^2,...,a^{d-1} \}$is a subgroup of$G$. By Lagrange, we have:$|G| \vdots |H|$, so$|G| \vdots d$; then, there is an$k \in \mathbb{N}$such that$kd=|G|$. From this and$a^d=e$we get that$a^{dk}=e^k$or$a^{|G|}=e$. 0 Yes, if$N$(for nontrivial) is any nontrivial group (for instance the one with two elements), then the projection$H=G\times N\to G$on the first factor has kernel$N\subseteq H$, so$G\cong H/N$. 0 If you want to be precise, you must specify what your$a,b,c$are.$\def\nn{\mathbb{N}}\def\zz{\mathbb{Z}}$Given$n \in \nn^+$and$a,b,c \in \zz/n\zz$, we have the exact equality$(a+b)+c = a+(b+c)$and$(ab)c = a(bc)$. Given$n \in \nn^+$and$a,b,c \in \zz$, we have an equivalence relation which we can denote by "$\equiv_{\pmod{n}}$" which lets us ... 1 Using the morphism $$\mathrm{sign} : x \mapsto \left\{ \begin{array}{cl} +1 & \text{if} \ x >0 \\ -1 & \text{if} \ x<0 \end{array} \right.,$$ you can notice that$\mathbb{Q}^*$(or$\mathbb{R}^*$) has a finite quotient, namely$\mathbb{Z}_2$. On the other hand,$\mathbb{Q}$(or$\mathbb{R}$) has no non-trivial finite quotient. Indeed, for any ... 5 Yes, for example$H:=G\times G$and$N:=G\times e$(if$G$is trivial just take any nontrivial group$H$and$H=N$, for example$H=N=\Bbb Z $) 0 In a direct product$G_1\times G_2$of two groups the subset$\{(g,e)\mid g\in G_1\}$is always a normal subgroup (isomorphic to$G_1$). Hence quotienting by it gives$G_2$. So any group (whatever be its order) can be obtained as$G/N$for suitable$G$and a suitable normal subgroup$N$there. 0$\Bbb Z_p$is a shorthand for$\Bbb Z/p\Bbb Z$. When$p$is prime this defines a quotient ring, since$\Bbb Z$is a ring and$p\Bbb Z$defines an ideal of$\Bbb Z$. It turns out that$p\Bbb Z$is a maximal ideal in$\Bbb Z$and hence the quotient ring$\Bbb Z_p$is a field, e.g.$\Bbb F_p$the field of$p$elements. Now what does$\Bbb Z/p\Bbb Z$mean in ... 6 The set$SL(2,\mathbb F)is not an abelian group, because $$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix}= \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}\ne \begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}= \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & ... 1 \mathbb C/\mathbb Z is isomorphic to \mathbb C/\mathbb n\mathbb Z for any non-zero complex number n. You can show that \mathbb C/\mathbb Z\cong (\mathbb C\setminus\{0\},\times) via the homomorphism z \to e^{2\pi iz}. You can also think of this as S^1\times\mathbb R, which is topologically a cylinder. 1 Presumably the question is referring to the other faithful representation of S_4 in three dimensions, namely the group of all symmetries of a regular tetrahedron. 2 Suppose that V is an inner product space. Let T:V \to V be a linear transformation. What is known as the adjoint of T, denoted by T^*, has the defining characteristic$$ \langle T(\textbf{v}),\textbf{w}\rangle=\langle\textbf{v},T^*(\textbf{w})\rangle , \mbox{ for all }\textbf{v},\textbf{w}\in V.$$Furthermore, T is referred to as unitary precisely ... 2 \mathbb Z is a free group. We can let S=\{1\} and then indeed have the definig property of dree group: For any group G and map f\colon S\to G there exists one and only one group homomoprhism \phi\colon\mathbb Z\to G with the property that \phi(x)=f(x) for all x\in S. This is just a highbrow formulation for the homomoprhism n\mapsto f(1)^n. ... 1 This is more of a "long(ish) comment" than an answer, and these are mostly my instincts speaking. But, I come bearing references! While I won't say there's a direct connection between this and the Catalan numbers, I do suspect there is a connection, particularly involving the generalized Catalan numbers which count the ways to subdivide regular n-gons ... 3 This problem is identical to counting non-crossing handshakes on a polygon, which is enumerated by the Catalan numbers C(n). In this case, the only difference is that all n vertices on the polygon are considered, while the n^{th} Catalan number corresponds to a 2n-gon.$$Q(n,2) = C(\frac{n}{2})$$This gives us some clues as to a recurrence relation ... 2 Put T_n = Q(n, 2) and taking T_0 = 0 by convention. Then the T_n are odd when n is odd and the values for even n satisfy the following recursion equations:$$ \begin{align*} T_0 &= 1 \\ T_2 &= 1 \\ T_n &= \sum_{i=1}^{n/2}T_{2(i-1)} \cdot T_{2(n/2-i)} \end{align*} $$To see this, enumerate the vertices of the n-gon as v_0, \ldots, ... 12 You can do this by contradition. Let's assume that (\Bbb Q, +) and (\Bbb Q^*, \cdot) are isomorphic. Then there exists a isomorphism \varphi: (\Bbb Q, +) \to (\Bbb Q^*, \cdot). Since \varphi has to be surjective, there exists a q \in \Bbb Q, such thtat \varphi(q) = -1. Now we calculate using the homomorphism properties of \varphi$$ -1 = ... 21 In the group of nonzero rationals with respect to multiplication there is an element which is its own inverse, namely-1$. But no such element exists in the group of rationals with respect to addition. Such elements must be preserved by any isomorphism. 0 I concur with Derek that answer in general is NO. However, since conjugate subgroups have the same order, you might try to find the normalizer of a subgroup$H$, that is$N_G(H)=\{g \in G| H^g=H \}$. Namely, index$[G:N_G(H)]$, equals the number of different conjugates of$H$. However, two subgroups of the same order do not have to be conjugate. 6 Hint A ring isomorphism is in particular an isomorphism of the underlying abelian group under addition, but the abelian groups$n \Bbb Z$,$n > 0$, each have only two automorphisms; using this one can show that there are only two isomorphisms (of abelian groups)$n \Bbb Z \to m \Bbb Z$for any$n, m > 0$. 8 In$2\mathbb Z$, there exists a generator$a$of the additive group such that$a+a=x^2$has a solution. This is not the case for$3\mathbb Z$. 0 Note that$\langle(1~3)\rangle=\{\epsilon,(1~3)\}$where$\epsilon$is the identity in$S_3$. Noting$(1~2)^{-1}=(1~2)$we have that $$(1~2)\langle(1~3)\rangle(1~2)^{-1}=\{(1~2)\epsilon(1~2),(1~2)(1~3)(1~2)\}=\{\epsilon,(2~3)\}=\langle(2~3)\rangle\neq\langle(1~3)\rangle$$ In particular$\langle(1~3)\rangle$is not a normal subgroup of$S_3\$. The subgroup ...

Top 50 recent answers are included