# Tag Info

8

The multiplicative groups ${\Bbb F}_3(X)^\times$ and $\Bbb Q^\times$ are both isomorphic to the sum of $C_2$ (corresponding to $\Bbb F_3^\times$ and $\Bbb Z^\times$) with a free abelian group of countable rank (generated by the primes of $\Bbb Z$ and the irreducible polynomials in $\Bbb F_3[X]$ respectively).

7

There are infinite groups which cannot be represented by finite-dimensional matrices over any commutative ring. If a group can be represented by matrices in such a way then it is called linear (perhaps this definition really requires field not commutative ring, but everything in this answer works for the more general commutative ring definition). Not all ...

6

The center of $G$ consists of those elements $g \in G$ such that $gh = hg$ for all $h \in H$. In an abelian group, this relation always holds, so $Z(G) = G$. It follows that $G / Z(G)$ is the trivial group, consisting of only a single element (namely, the coset $Z(G)$). Every group with one element is cyclic.

6

A partial result: there exists a unique group of order $n$ (i.e. the cyclic group) if and only if $(n,\phi(n)) = 1$, where $\phi$ is the Euler phi function and $(a,b) = \operatorname{gcd}(a,b)$. This is certainly satisfied when $n$ is a prime, but when $n = 15$ we have $\phi(n) = 8$, and since $(15,8) = 1$ there is a unique group of order $15$, even though ...

6

Try the map $aH \mapsto x(aH)x^{-1}$. Note that $x(aH)x^{-1} = (xax^{-1})xHx^{-1}$. More generally, if $\phi$ is an automorphism of $G$ and $H \leq G$, then $[G:H] = [G : \phi(H)]$ since $aH \mapsto \phi(aH)$ is a bijection between left cosets of $H$ and $\phi(H)$. Your problem is the case where $\phi(g) = xgx^{-1}$.

5

You mean like this? http://ruwix.com/online-rubiks-cube-solver-program/solution.php?cube=0343515641165422615412533412316442361454656232126363525&x=1 To find the solution just click "play". EDIT: By the way, the movements to get this setup are: L R' F' B M' E' U B B Anyway the question is quite interesting because I think there are not tons of ...

5

Start with any nonidentity element $g \in G$. We see that the cyclic subgroup $\langle g \rangle$ generated by $g$ is in fact equal to $G$, so $G$ is cyclic. If $G$ were infinite, we would reach a contradiction. Knowing that $G$ is finite, we have $G \cong \mathbb{Z}/n\mathbb{Z}$ for some integer $n$. What if $n$ were not prime?

5

Let $G$ be a group. The order of an element $g \in G$ is the smallest $n \in {\mathbb N}, n \geq 1$ such that $g^n = 1$. If a group is cyclic, then there is an element $c$ such that $\{1, c, c^2, \dots, \}$ is the whole group. In the case that this set is finite, you have $c^n = 1$ for some $n \geq 1$ (and after that the sequence repeats); the least such ...

4

There are two basic ideas: Given $X$, if it is finite then pick any bijection with $\Bbb Z/(n)$, and you have a finite group; otherwise consider $\Bbb Z[X]$, the ring of polynomials whose free variables are elements of the set $X$. We can prove, using the axiom of choice, that $\Bbb Z[X]$ has the same cardinality as $X$. Therefore there exists a bijection ...

4

The parity of an element $sgn(\sigma)$ gives you a linear (irreducible) character (one-dimensional representation). Hence, by Frobenius orthogonality relations, the inner product with the other characters is $0$, and this is exactly your sum! (Note that conjugate permutations have the same sign)

4

The inclusion in $(1)$ cannot be an equality for every $i$, in general. If $G$ is nilpotent, then $G = \zeta^i(G)$ for some $i$, and also $[G, G] \subsetneq G$, so $[\zeta^{i+1}(G), G] \subseteq [G, G] \subsetneq \zeta^i(G)$. A nice way to see $(2)$ has already been addressed in the comments above.

4

Usually the way to approach matrix polynomial equations (of one variable) is to figure out what the minimal and/or characteristic polynomial of the solution matrices would be. Note that any conjugate of a solution to $p(A)=0$ is also a solution, so if there is one solution there will be infinitely many (except in cases where the only solutions are scalars). ...

3

Let $S=(12)$ and let $T = (13)$. Then $\langle (12)\rangle \cup\langle (13)\rangle$ has three elements (And is not even a group!). But, $\langle (12), (13)\rangle = S_3$. This has 6 elements and IS a group. You can do something quite similar for the intersection. The elements $S$ and $T$ should actually work as a counter example.. Explicitly: ...

3

One could equally ask for the probability with ordered or unordered pairs, with different results. It suffices to count the number of pairs $a,b$ satisfying $x=ab\vee x=ba$, where $x\in G$ is fixed. Say we wish to count ordered pairs of not necessarily distinct elements. Notice the equivalence $x=ab\vee x=ba\iff b=a^{-1}x\vee b=xa^{-1}$. If we naively ...

3

As vadim123 mentioned in his comment, there is always a trivial order. Suppose you have some non-trivial order that respects the group structure on your group $G$. Suppose $g\in G$ and $1\leq g$. Then multiplying by $g$ you will get $g\leq g^2$, $g^2\leq g^3$ et.c. But since $G$ is finite, eventually $g^n=1$ for some $n$. Then you have $1\leq g\leq g^2\leq ... 3 You can't embed$\omega_1$into the$\Bbb{R}^n$or$\Bbb{N}^n$(in the sense of topological embedding.) Because if$A$is a subset of$\Bbb{R}^n$then it has only countably many isolated point, but$\omega_1$has$\aleph_1$many isolated points. For the same reason, you cannot embed$\omega_1$into$\Bbb{N}^n$, even you cannot embed it into$\Bbb{R}^\omega$... 3 The order of an element is the order of the cyclic group generated by the element. I believe it would be more correct to say the period of an element, but to say order i common. In your case$C_2\times C_2=\{e,a,b,ab\}$with$a^2=b^2=(ab)^2=e$. So the order (more correctly: period) of$a$is 2 as the cyclic group generated by$s=a$has two elements. If ... 3 A representation of a group$G$is a morphism$\rho:G\rightarrow\text{GL}(V)$where$V$is a vector space over a field$K$. There are always at least the following two representations: the trivial representation, where$\rho(g)=\text{id}_V$for all$g\in G$. the regular representation, where$V$is the space of$K$-valued functions on$G$and ... 3 The stipulations$A\subseteq C$and$A\cap B=\varnothing$tell us that$A\subseteq C\setminus B$. The stipulation$\langle A\rangle=C$tells us that$A$has to be "big enough" (enough to generate all of$C$). Might as well pick$A=C\setminus B$. To show$\langle C\setminus B\rangle=C$it suffices to show$B\subseteq\langle C\setminus B\rangle$. Can you show ... 3 Here are two possibilities for$n$for which there are precisely two groups of order$n$.$n=p_1p_2 \cdots p_n$for some$n \ge 2$and distinct primes$p_i$, such that there is exactly one pair$(i,j)$with$p_i$divides$p_j-1$.$n=p_1^2p_2\cdots p_n$for some$n \ge 1$and distinct primes$p_i$, where there are no$(i,j)$with$p_i$divides$p_j-1$, and ... 3 By the earlier answers, all of the elements of order$4$all have minimal polynomial$x^2+1$, so you can assume that one of them is$A = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$. Suppose$B = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$is another matrix of order$4$in the image. Then, from the structure of$Q_8$, we have ... 3 You can consider the obvious surjective homomorphism$G_1\times G_2\to (G_1/H_1)\times (G_2/H_2)$. What is its kernel? Then you can use the first isomorphism theorem. This is essentially the same solution as if you just defined a map and checked that it is an bijective homomorphism. But using the 1st isom. theorem you don't need to do this many ... 2 (1): Consider$G = D_8 \times C_2$, where$D_8$is dihedral of order$8$and$C_2$is cyclic of order$2$. For this group,$\zeta^2 = G$and$[G,G]$is a proper subgroup of$\zeta^1$. (2): If$N$is characteristic in$G$, then we have the map$\operatorname{Aut}(G) \rightarrow \operatorname{Aut}(G/N)$defined by$\phi \mapsto \hat{\phi}$, where ... 2 The statement says that$G'$equals the normal closure N of the set of commutators of the elements of$S$, which is the smallest normal subgroup containing the set of such commutators. To prove this, divide G by N. The quotient group is abelian since it's generators commute. Thus,$N$contains$G'$. It is also clearly contained in$G'$. Hence, they are ... 2 I have thought of one way of doing this, but there might be easier ways. Let$\mathcal{F}$be a functor with the properties you describe. Suppose that we have groups$G$,$H$with homomorphisms$\phi:G \to H$and$\psi:H \to G$with$\psi \phi = {\rm Id}_G$. Then$\mathcal{F}(\psi\phi) = \mathcal{F}(\psi) \mathcal{F}(\phi)$is the identity map on ... 2 I will summarize the proof from the book "Theory of Groups of Finite Order" by W. Burnside. It would not be very easy to think of this if you had not seen it before! With your notation, conjugation by$a$is inducing a fixed-point-free automorphism of order$3$of$G$. You know that$x$commutes with$a^{-1}xa$and similarly with$axa^{-1}$. If you ... 2 Hint: For showing$\epsilon(\sigma_1\sigma_2)\equiv\epsilon(\sigma_1)+\epsilon(\sigma_1)$mod$2$where$\sigma_i\in S_n$for$i=1,2$First, show that$\epsilon(\sigma\lambda)=\epsilon(\sigma)\pm1$where$\lambda$is transposition. Second, A permutation cannot be written as a product of both an odd and an even number of transpositions. 2 You have a normal subgroup of order$5$. Your calculations are already sufficient to show that the elements of this group other than the identity don't commute with$\tau$, or indeed any of its powers. So$\tau, \tau^2, \tau^3$are not in the centre.$1=\tau^0=\tau^4$is of course in the centre. Suppose we have an element$\rho$which is in the centre, and ... 2 There is a theorem of Ito: let$q$be a prime power,$d$a positive integer, and$p$a prime divisor of$q−1$with$d \leq p$. Fix some field$F$of order$q$and some element$\alpha$of this field of order$p$. The Frobenius complement$H$is the cyclic subgroup generated by the diagonal matrix whose$(i,i)$th entry is$\alpha^i$. The Frobenius kernel$K$... 2 Suppose$K$is a finite group with a fixed-point-free automorphism of prime order. Then Thompson (1959) showed$K$is nilpotent. Suppose$K$is a finite solvable (or locally nilpotent) group with a fixed-point-free automorphism of prime order$p$. Then Higman (1957) showed$K$is nilpotent with nilpotency class bounded above by a function,$k(p)$, of$p\$. ...

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