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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 ...



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