# Tag Info

1

$$e^x\mapsto (x,0)$$ $$-e^x\mapsto (x,1)$$ Gives the isomorphism.

1

$\mathbb{R/Z}$ is isomorphic to the unit circle $S^1$, the set of all complex numbers of absolute value $1$, seen as a multiplicative group. $\mathbb{Q/Z}$ is isomorphic to the torsion subgroup of $S^1$, formed by the elements of finite order, that is, all roots of unity. $\mathbb{R/Q}$ is thus isomorphic to $S^1/tor(S^1)$, a very large and complicated ...

1

the Galois group $Gal(\Bbb{Q}(\zeta_p))$ is isomorphic to $(\Bbb{Z}/p\Bbb{Z})^*,\times)$ this also isomorphic to $(\Bbb{Z}/(p-1)\Bbb{Z}),+)$, and this last group is cyclic of degree $(p-1)$, so for every divisor $m$ of $(p-1)$, there are only an sub group of $(\Bbb{Z}/(p-1)\Bbb{Z}),+)$ of order $m$ . The hypotheses that $3$ divide $p-1$ give only an sub ...

0

for each $p$ a prime number we get a sub group $H_p=\{\left(\begin{array}{cc} 1 & d \\ 0 & \frac mn \end{array}\right) ,\mid d\in\Bbb{C}, (n,p)=1,(m,p)=1, n,m\in\Bbb{Z}^*\}$, $H_p$ is a subgroup because is not empty the identity is an element, stable by product, and each element have an inverse in $H_c$ , and is normal in $G$ because ...

0

Let $g$ and $h$ be two elements in $G$. Calculate $ghg^{-1}$. You will immediately see under what conditions $ghg^{-1}$ is "of the same form" as $h$. That is, $ghg^{-1} \in H$ where $H$ is a subgroup of $G$. This condition $ghg^{-1} \in H$ means that $H$ is normal. This method will give you infinitely many subgroups.

0

The extension is actually degree 4 and is easily seen to be Galois. You just must show that $\tau$ has order 4, which by Lagrange really just amounts to showing that it has order greater than 2 and it will generate the whole group. So see what $\tau^2$ does on your generator and observe it is not the identity and the claim will follow.

0

Hint: For the first question, to describe an element of $\operatorname{Gal}(L/{\bf Q})$, you only need to say where $\sqrt[4]{5}$ goes and where $i$ goes, and check that the choice leads to an automorphism. For the second question, just use the Galois correspondence.

1

A subgroup $H \subset G$ is normal in $G$ if and only if $H$ is the kernel of a homomorphism, if and only if the operation $aH*bH:=abH$ is well defined, if and only if for all $a, b \in G$ $aHbH:=\{ah_1bh_2 | h_1, h_2 \in H\}=abH.$ Amongst other things, this last claim follows from the fact that $$ah_1bh_2=ab(b^{-1}h_1b)h_2,$$ and now the expression in ...

0

Here is another answer using the standard action of $S_4$ on the set $\{1,2,3,4\}$. We use the following lemma (proof below) Lemma: If $(C_2)^3$ acts on a set $X$ with four elements, then there is at least one nontrivial element of $(C_2)^3$ which acts trivially. If $(C_2)^3$ were a subgroup of $S_4$, then $(C_2)^3$ would act on $\{1,2,3,4\}$ by ...

0

Yes, is true. I f $\;x\in C\;$ , then any element in $\;C\;$ is of the form $\;x^g:=g^{-1}xg\;$, for some $\;g\in G\;$ , and if $\;x\in N\lhd G\;$ , then also $\;x^g\in N\;$ . From here you could also characterize subgroups that are normal as those subgroups which are the (disjoint, of course) union of conjugacy classes in the group

2

Suppose $G_{n+1}/G_n$ is not simple, then we have $NG_n$ an intermediate normal subgroup. Rinse and repeat.

1

Into the last factor, we should also place the largest power of each prime for similar reasons, which in this case is $C_9$ and one copy of $C_5$, $C_7$ and $C_4$ since if we didn't then we break the condition that $n_i|n_{i+1}$. For example, if we put in $C_3$ instead of $C_9$, then we would earlier get that $9$ divided $3$ which is wrong. This gives the ...

1

You can write for example: $$4\cdot5\cdot7\cdot9\;,\;\;\text{and}\;\;\;3\cdot5\;\;\implies G\cong C_{15}\times C_{4\cdot5\cdot7\cdot9}$$

3

You have already noted that $$C_{12}\times C_{35}\times C_{45}=C_3\times C_4\times C_5\times C_5\times C_7\times C_9.$$ Now for each prime $p$ take the product of the $C_{p^k}$ with $k$ maximal, without repeats. So in this case we get $$C_4\times C_5\times C_7\times C_9\cong C_{4\times5\times7\times9}=C_{1260}.$$ Now repeat the process with the remaining ...

1

in group theory, I think better to say, a subset S of group G is called free if any strict subset A of S we have $\langle A\rangle \varsubsetneq \langle S\rangle$

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Given the paper mentioned by the OP and one of its references (with the exact same title—ugh!)... Given $\phi\in\operatorname{Aut}(G)$ we say that $\phi$ centralizes a subgroup $H\subseteq G$ if $\phi(h)=h$ for all $h\in H$. So $\phi$ fixes $H$ elementwise. Equivalently, the restriction map gives a group homomorphism ...

1

Hint: $G = \langle (135),(27),(468)\rangle$ and $N = \langle (27),(468)\rangle$.

2

Your group is abelian becaus $\sigma$ and $\tau$ commute, the 2-Sylow is the same to the 2-sylow of $\langle\sigma \rangle$ and $\langle\tau\rangle$ because $\sigma^3=\tau^3$, also your groupe have 3-sylow isomorphic to $Z_3\times Z_3$ and then $G/\langle\tau\rangle$ is $G/\langle\sigma\rangle$ isom $Z_3$

2

A basis does not have to be Nielsen reduced: $(x^{-1},xy)$ is a basis for the free group $\langle x,y \rangle$, but $|x^{-1}xy|=|y|<2=|xy|$ (which contradicts one of the requirements to be Nielsen reduced). It is true that you can reduce finite sets to Nielsen reduced set, by using Nielsen transformations, and this is discussed in Combinatorial Group ...

2

As already noted in a comment, $p_1$ is already exhibited as a product of disjoint cycles, in this case one cycle. A product of one factor is, by the way, not the extreme case of a product; there is also the empty product, the identity element. Whereas evaluating the empty product requires you to know the identity element, a product with a single factor ...

3

By definition, a simple module is nonzero. Hence its dimension, which equals its multiplicity in the decomposition, is positive. This guarantees that any irreducible representation does appear in it.

0

A theorem of Brauer says that the number of simple $kG$-modules (up to isomorphism) is equal to the number of conjugacy classes of $p$-regular elements of $G$ ($g \in G$ being $p$-regular means the order of $g$ is not divisible by $p$). If $G$ is not a $p$-group, there are at least two such conjugacy classes. Thus there are at least two simple ...

1

Hint: If $x \in H_i \cap H_j$ ($i < j$), then $x \in H_i \implies x \in H_1 H_2 \cdots H_{j-1}$. Solution: Let $x \in H_i \cap H_j$, for any $i \ne j$. Without loss of generality, let $i < j$. Then, $x \in H_i \implies x \in H_1 H_2 \cdots H_{j-1}$, which further implies that $x \in H_1 H_2 \cdots H_{j-1} \cap H_j$. But $H_1 H_2 \cdots H_{j-1} \cap ... 1 You are almost there! Let$S$be a subgroup of$S_4$of order$8$. The orbit of$S$under the conjugation of$S_4$has size$24/|Stab(S)|$. Now note that the stabilizer of$S$for this action is$N(S)$, the normalizer of$S$in$S_4$(i.e., the biggest subgroup of$S_4$in which$S$is normal). In particular$N(S)$has size$8$or$24$since it contains$S$, ... 2 I know one example of this, but it might not be the smallest. Let$N = {\rm SL}(2,9)$, which is isomorphic to a double cover$2.A_6$of$A_6$. Then${\rm Out}(N) \cong C_2 \times C_2$. Let$G = {\rm Out}(N)$with$\chi$the identity map. Then there is no extension$E$that induces this coupling. In the ATLAS of Finite Simple groups, if you look under ... 4 The infinite product${\mathbb S}^1 \times {\mathbb S}^1 \times ...$satisfies this requirement. It is abelian, compact and connected, but it is not a Lie group because it has an infinite, strictly increasing chain of connected, closed subgroups. 4 Clearly$o(n) = o(n^{-1})$so this means there must be exactly one element of order$2$. If you mean this for all$n$, then this means the group must just be$\Bbb Z/2\Bbb Z$which is isomorphic to$S_2$. So$1$,$3$,$4$all hold. On the other hand if you just mean this to be true for some$n$, I must think you're crazy since again we see that$n$is$2$, ... 1 1) Check that a subgroup$\;H\;$of a group is a normal subgroup iff it is the (disjoint) union of conjugacy classes. 2) With (1), or directly, check that$\;V:=\left\{\,(1),\,(12)(34),\,(13)(24)\,,\,(14)(23)\,\right\}\lhd S_4\;$3) Take now the quotient group$\;S_4/V\;$. By Lagrange's theorem, this group's order is six, so it is either$\;S_3\;$or the ... 1 Consider the identity map$id: V \to V$. Since$id$is bijective and commutes with the$G$-action, it is an isomorphism of representations. This implies that for any decomposition $$V= \bigoplus_j W_j^{\oplus b_j}$$ it must be that for each$W_j^{\oplus b_j}$there exists$V_i^{\oplus a_i}$such that$id(W_j^{\oplus b_j})=V_i^{\oplus a_i}$, otherwise ... 0 A normal subgroup, as you noted, is a union of conjugacy classes. It must also contain the identity element, which accounts for$1$element of$H$. Do the sizes of any of the other conjugacy classes plus$1$give you$4$? Edit: Now that you know the correct subgroup, here's something to think about for proving$S_4/H\simeq S_3$. Note that every element of ... 2 By the Schur-Zassenhaus theorem, every extension $$1\rightarrow M\rightarrow E\rightarrow G\rightarrow 1$$ with$gcd(|M|,|G|)=1$is split, i.e.,$E\cong M\rtimes G$. In other words, we have$H^2(G,M)=1$. We can apply this here with$G=Aff(q)/H\cong \mathbb{F}_q^*$of order$q-1$and$M=H\cong \mathbb{F}_q$of order$q$, because$gcd(q,q-1)=1$. So there ... 1 Since$Z$lies in the centre of$P$, certainly$P\le N_G(Z)$. 5 Homomorphism part is correct. However, you can't use the logarithm on complex numbers, as it is a multivalued function. Injectivity Part It is enough to check the kernel of the homomorphism is trivial.$e^{in\theta}=1$implies$n\theta=2k\pi$for some$k\in\mathbb{Z}$. So, the map is injective iff$\theta$is not a rational multiple of$\pi$. 0 Inspired by Mariano Suárez-Alvarez♦'s: Just like Mariano Suárez-Alvarez♦ comments: Given me a non-prime field$K$.$K$has some proper subfield$L$and$K$can be viewed as an$L$-vector space. Every vector space has a basis. Say$\{a_{i}:i\in I\}$is a basis of$K$over$L$. Consider a map$f:\{a_{1},a_{2}\}\rightarrow \mathbb{K},~a_{1}\mapsto 0, ...

4

If the field is a prime field (one of the $F_p$ with p prime, or $\mathbb Q$) then you can't, and this for exactly the reason you mention (which works also for the rationals). If not, you can. Indeed, in that case the field K contains a proper subfield L, and you can find an L-linear map which is not K-linear.

0

The first implication is right but you lost a sense of proriété The second implication is false, which allows you to recover the lost logical property the result is true. So write better: $xHx^{- 1} H = H \forall x \in G$ then $xhx^{- 1} H = H \forall x \in G$ and $\forall h \in H$ then $xhx^{- 1} \in H \forall x \in G$ and $\forall h \in H$ therefore H ...

3

Is this correct? Yes. Infinite Galois theory provides a bijection between subfields of $\overline{\mathbb Q}$ and closed subgroups of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Since every open subgroup of a topological group is closed, and the open subgroups all have finite index in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, both your ...

2

Note that $x^n=e\iff (x^{-1})^n=e$.

1

If $H$ is polynormal and $H \le K \unlhd L$ then, for $l \in L$, $H^l \le H^{\langle l \rangle} = H^{H^{\langle l \rangle}} \le H^K$. Conversely, $H^{\langle x \rangle} \unlhd \langle H,x \rangle$ so, assuming the condition, $H^{H^{x}} = H^{\langle H,x \rangle} = H^{\langle x \rangle}$.

0

This is known as the kernel pair of a morphism.

0

$(abc)$ maps $a$ to $b$, $b$ to $c$, and $c$ to $a$. Now, all we need to do is map $c$ to $a$ (so $b \mapsto c \mapsto a$), $a$ to $d$ ($c \mapsto a \mapsto d$), and $d$ to $c$ — i.e., $(cad)$. Thus, $(ab)(cd) = (cad)(abc)$.

2

Try $(ab)(cd) = (ab)(bc)(bc)(cd)= (abc)(bcd)$.

0

Obviously, a characteristic subgroup $H \subseteq G$ is strongly potentially characteristic. Here is an example of a non-characteristic subgroup. Consider the groups: $$H = \{ 0 \} \times \mathbb{Z}/2 \subset \mathbb{Z}/2 \times \mathbb{Z}/2 = G$$ The subgroup $H$ is not characteristic in $G$, because the automorphism of $G$ switching the coordinates does ...

4

The multiplicative group of matrices $$\left\{ \begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix} \middle\vert\, a,b,c \in \mathbb{F}_3\right\} \subset \operatorname{Mat}(3 \times 3, \mathbb{F}_3)$$ is a counterexample (it is isomorphic to the group in Joanpemo's answer).

0

If $G$ is finite it is equivalent to say that $f$ is injective, i.e. there are no elements of order $2$. If $G$ is finitely generated abelian then it must be finite (bt the structure theorem on these groups) and the previous statement applies. In general I don't know of any characterization. For example, take a field $K$ (here the group is $K$ with ...

4

Nope. The (non-trivial) semidirect product $\;C_3\ltimes(C_3\times C_3)\;$ has exponent $\;3\;$ and it is certainly non-abelian.

1

Let $\Sigma_{g}^{1}$ be denote a surface of genus $g$ with one boundary circle. Let $\iota_g\colon \textrm{S}^{1} \to \partial\Sigma_{g}^{1}$ be a homeomorphism on the boundary. For convenience later on, we write $\textrm{S}^{1} = [0,1]/\{0 \sim 1\}$, and we fix base points $x_g = \iota_g(0) \in \Sigma_{g}^{1}$. I assume that the following statement is ...

2

1) {1,-1}. Identity is 1, $-1^{-1}=-1$ Can prove it is the only one by considering if $b$ is in the group so are all infinite $b^n$. 2) let $b$ be any real number not equal to 1 or -1. Then {$b^n$, *} is an infinite group. 1 is the identity and $b^n*b^m=b^{m+n}$. $(b^n)^{-1}=b^{-n}$. 3) this can't be done. If $b$ in the group then so is ...

2

The commutator subgroup of a group $G$ enjoys the property of being the smallest normal subgroup of $G$ with abelian quotient. This often helps when you are trying to find the commutator subgroup of a given group. For example, if you can get your hands on all the normal subgroups, you can go calculating quotient groups, starting with the smallest normal ...

1

Since $\;|G|=p^a\;$ , we have that for any $$\;x\in G\;,\;\;x^{p^a}=1\implies \text{ord}\,(x)\,\mid\,p^a\,\mid\,p^k\implies\text{ord}\,(x)\,\mid\,p^k\;,\;\;\text{for}\;\;a\le k$$ Fill in details now and complete the argument.

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