# Tag Info

### Find the smallest $n$ such that $S_n$ has an element of given order.

Hint $\lvert g\rvert =\rm {lcm}(|c_1\rvert, \lvert c_2\rvert, \dots, \lvert c_k\rvert)$, where $g=c_1c_2\dots c_k$ is the cycle decomposition of $g$. Write $$m=\prod_ip_i^{a_i}$$, for the prime ...

### split maximal torus construction

The subgroup $T$ is the image of the subgroup of diagonal matrices in $GL(n,q)$. MAGMA constructs $PGL(n,q)$ as a permutation group, acting on the projective line which has $(q^n - 1)/(q-1)$ points. ...
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For any field $\mathbb{F}$, the symplectic group $Sp(2n,\mathbb{F})$ is generated by symplectic transvections. These are maps of the form $f_{\alpha,u}$, where $\alpha \in \mathbb{F}$, $u \in \mathbb{... 1 vote Accepted ### Let$H,K$be subgroups of$G$. Show that if$G$has elements$x,y$such that$xH=yK$, then$H=K$. Is this a correct reasoning? No, as you've written it doesn't make sense, because$y^{-1}x\in G$doesn't mean that$h\in K$. If$xH=yK$then$xe=x=yq$for some$q\in K$, so$y^{-1}x=q\in K$. Once you'... 0 votes Accepted ### Abstract Algebra Normal Subgroups This just comes down to basic coset manipulations. You have $$h_1K = \{ h_1 k : k \in K\}.$$ You always have$h_1 \in h_1K$. Therefore, if$h_1K = h_2K$, this means that there exists a$k \in K... 2 votes ### Isomorphism between two presentations of 6 order abelian group One can use von Dyck's Theorem to obtain the morphisms; this will save you the work of checking the maps are homomorphisms. Let \begin{align} G&= \langle k\mid k^6\rangle,\\ K&= \langle s,r\... 1 vote ### Outer automorphism group being a quotient group Here's another perspective on Out(G) that might be clearer: Consider the map \phi:G\to Aut(G) by \phi(g)= \phi_g such that \phi_g(h)= g^{-1}hg for all h\in G. It's a good exercise to prove ... 1 vote Accepted ### Outer automorphism group being a quotient group \text{Out}(G) is supposed to measure the "outer" automorphisms of G. That is, the automorphisms that aren't inner. To do this, we start with the group of all automorphisms \text{Aut}(G)... 3 votes ### Isomorphism between two presentations of 6 order abelian group I think to have found it explicitly. Let's redefine the two groups like this:A=\langle r\mid r^6\rangle \\ B=\langle a, b\mid a^3, b^2, aba^{-1}b\rangle $$We can define a homomorphism \Phi = A\... 3 votes ### Application of nonfamous finite groups in computer science Automorphism groups of codes. For example, the sporadic Mathieu group M_{24} is the automorphism group of the extended binary Golay code. 2 votes Accepted ### Application of nonfamous finite groups in computer science Isomorphism testing is the area that comes to mind for me. In Graph Isomorphism (GI), the key techniques are Weisfeiler--Leman (WL) and Permutation Group Algorithms. At a high level, Babai uses WL to ... 0 votes ### Wilson's proof of Iwasawa's Lemma A point-stabilizer subgroup depends on a choice of point. However \mathrm{Stab}(gx)=g\mathrm{Stab}(x)g^{-1} implies all points in an orbit have conjugate stabilizers. Thus, we can speak of "the&... 6 votes ### Does there exist a group homomorphism (\mathbb{Q}_p, +) \to (\mathbb{R}, +)? As mentioned in the comments, as \mathbb Q-vector spaces (and hence also as abelian groups), \mathbb Q_p\cong\mathbb R. Thus, there are many group homomorphisms between them. This "morally&... 2 votes Accepted ### Are the roots of unity the only algebraic subgroups of the multiplicative group? Case of fields Let me begin by confirming your suspicion over a field. Claim (field case): Let k be a field, and H\subsetneq \mathbf{G}_{m,k}. Then, H is isomorphic to \mu_{n,k} for some n\... 1 vote Accepted ### Does a free group F of finite rank n have finitely many retracts (as a subgroup)? I close this question by giving the answer of @MoisheKohan in the comments: For n\geq 2, the statement is false because each free factor of F_n is a retract and there are infinitely many free ... 3 votes Accepted ### Why is addition on elliptic curves defined in this particular way? The group law on elliptic curves comes from looking at the functions of geometric surfaces, and trying to understand the behaviour of intersections. When we look at the coordinate ring of an elliptic ... 3 votes Accepted ### Are metacyclic p-groups semidirect products? It turned out to be quite difficult to find a good link to answer this question. Therefore I will give here the full formulation of the theorem on metacyclic p-groups. Theorem. For odd p, every ... 2 votes ### Prove that S is a coset of some subgroup of G iff S+S-S=S. This is @AtticusStonestrom's proof in more detail. Since \varnothing\neq S, let t\in S. It suffices to show that H:=S-t=\{ s-t\mid s\in S\} is a subgroup of G, since then S=H+t is a coset of ... 4 votes Accepted ### Prove that S is a coset of some subgroup of G iff S+S-S=S. Since S is non-empty, pick any t\in S. I claim that S-t:=\{s-t:s\in S\} is a group, which will show the desired result. Clearly 0\in S-t. Thus we need only show that, for any x,y\in S-t, we ... 1 vote Accepted ### Question about a proof for showing that A_n has no subgroup of order \frac{n!}{4} if n>4 Here \sigma is an arbitrary three cycle. The proof shows that \sigma \in H. Thus each three cycle is in H. But H cannot be A_n since H<A_n by definition. 2 votes Accepted ### Can there be a different proof be given for: If N is a subgroup of a group G of index 2, then N is a normal subgroup of G Since [G:N]=2, there is exactly two cosets of N in G. One of them must be eN=N. We have$$n\in N\iff nN=N.$$(Can you prove this?) This is equivalent to$$a\notin N\iff aN\neq N.$$If b\neq a... 1 vote ### If G_1\cong G_2, H_1\triangleleft G_1, H_2 \triangleleft G_2 and G_1/H_1\cong G_2/H_2, then is H_1\cong H_2? The third condition is NOT correct. Consider D_8, the dihedral group of order 8, which represents the symmetries of a square in the plane. A presentation for D_8 is$$D_8 = \langle r, s ~|~ r^4=1, ... 2 votes Accepted ### Quick question about a proof of the theorem: IfN$is a subgroup of a group$G$of index$2$, then$N$is a normal subgroup of$G$Let$a\in G. The function \begin{align} \lambda_a:N&\to aN,\\ n&\mapsto an \end{align} has inverse \begin{align} \lambda_{a^{-1}}:aN&\to N,\\ n&\mapsto a^{-1}n. \end{align} It ... 2 votes ### Quick question about a proof of the theorem: IfN$is a subgroup of a group$G$of index$2$, then$N$is a normal subgroup of$G$There's a natural bijection between the elements of$N$and the elements of$aN$given by$\phi(g)= a\cdot g$where$(\cdot)$is the group multiplication. It's a good exercise to prove this map is ... 1 vote Accepted ### Power of$ 2 $congruent to 1 mod n Ok so just summarizing what everyone else said, the solutions are exactly the multiples of$ o_n(2) $. So the number of solutions less than$ n $is the floor function of$ n /o_2(n) $. When 2 is ... 3 votes Accepted ### Determination of order of cosets in a factor group of a finite abelian group. Let$G_1=\Bbb{Z}_{3^{10}} $,$G_2=\Bbb{Z}_{3^{7}}$,$H=\langle (3^2, 3^3)\rangle$Then$|3^2|_{G_{1}}=3^8$,$|3^3|_{G_{2}}=3^4$Let$|a|=n$where$a=(1,0)+H$Then$n$is the least positive integer ... 1 vote ### First Isomorphism Theorem: Does each homomorphism has to be surjective? Is it possible to define an homomorphism$\phi:G\to H$such that$|G|<|H|$? Another standard homomorphism which is not surjective (for$|G|\ge 3$), is Cayley's one$a\mapsto(g\mapsto ag)$from$G$into$S_G$. 2 votes Accepted ### First Isomorphism Theorem: Does each homomorphism has to be surjective? Is it possible to define an homomorphism$\phi:G\to H$such that$|G|<|H|$? Homomorphisms can be injective, surjective, isomorphic, or neither of them. It only has to be invariant on the structure:$\varphi(a\cdot b)=\varphi(a)\cdot \varphi(b)$in case of (multiplicatively ... 5 votes ### Let$G$be a group with$25$elements and$E$a$G$-set with$32$elements. Show that there exists$a \in E$such that$G_a=G$. The orbits partition. Since the stabilisers orders have to divide$25$(they're subgroups), they're all of order$1,5$or$25$. So the same can be said for the orders of the orbits (orbit-stabilizer ... 4 votes Accepted ### Group action for signal For clarity, it should be made clear that$\mathcal{X}(\Omega)$is not a signal, but the vector space of all signals$x\colon\Omega\to\mathcal{C}$to some fixed space$\mathcal{C}$which is suppressed ... 4 votes Accepted ### Can we come up with a disjoint union of a subsets of the group$\Bbb{Z}$such that they do not equal the cosets of a subgroup, yet they form a group? It is not possible in any group. So, to be clear, we have a group$G$and we are trying to come up wit a partition$G=\sqcup_i A_i$, where "termwise multiplication" yields to a group ... 4 votes Accepted ### Why should I expect the generators of Lie Groups to be closed under the commutator? So the proper formal answer should be that the vector fields on a Lie group have a natural Lie bracket (as do all vector fields on a general manifold). Then the Lie algebra can be identified with the ... 2 votes ### Why is$\Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z$not isomorphic to$\Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z? We have that the first group is isomorphic to $$\Bbb Z_4\times(\Bbb Z_3\times \Bbb Z_4)\times (\Bbb Z_5\times \Bbb Z_8),$$ while the second group is isomorphic to $$\Bbb Z_8\times(\Bbb Z_2\times \Bbb ... 5 votes ### Why is \Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z not isomorphic to \Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z? One simple way to see this is that the second group has an element (0,5,0) of order 2 and no element g such that g+g=(0,5,0). There is no element like this in the first group. If (x,y,z)\in\... 1 vote Accepted ### Is there a way to define other homorphisms, different from the conjugation mapping, in the definition of outer semidirect-products? If (A,+) is an abelian group then the only group conjugation is the identity, but a\mapsto -a is an automorphism, and it is only the identity if every element of A has a+a=0. If (A,+) is ... 1 vote Accepted ### Writing a product of commensurable subgroups as a disjoint union Let g_1,\ldots,g_n be right coset representatives of H_1 \cap gH_2g^{-1} in gH_2g^{-1}. Then every element gh_2g^{-1} of gH_2g^{-1} has the form hg_i for some h \in H_1 \cap gH_2g^{-1} ... 0 votes ### Proving that a subgroup of a finitely generated abelian group is finitely generated Here is a proof by induction on number of generators. For n=1 it is easy to see that subgroup of a cyclic group is always cyclic. Now assume that the statement is true for all abelian groups with ... 1 vote ### Definition of the Zappa–Szép product of groups in categorical terms A definition by a universal property defines an object uniquely up to canonical isomorphism, so it cannot exist unless additional data is specified (the group is not defined by the subgroups ... 2 votes Accepted ### Inverse of 3 in multiplicative group C_{20}. I am going to answer of your original question. Let U(20) be the multiplicative group of C_{20}. We have 3\in U(20). What is the inverse of 3 in U(20) ? Suppose 3^{-1}=x. Then 3x\equiv ... 1 vote Accepted ### Express \mathbb{Z}^2/B as a direct product of cyclic groups Observe that \Bbb Z\cong 2\Bbb Z\cong 3\Bbb Z. Let G=\langle a,b\mid ab=ba\rangle\cong \Bbb Z^2. Suppose$$\begin{align} B&:=\langle a^2,b^2\rangle_G \\ &\cong 2\Bbb Z\times 2\Bbb Z\\ &... 1 vote Accepted ### Point of confusions for:H$and$K$are unequal subgroups of a group$G$, each of order$16$. Prove that$24 \leq |H \cup K| \leq 31$It is given that order of$H$and$K$both are$16$. Again$H \cap K$is a subgroup and as it is contained in both$H$and$K$. So$|H \cap K|$will divide$|H|$and as well as$|K|$. So we have ... 1 vote ### Point of confusions for:$H$and$K$are unequal subgroups of a group$G$, each of order$16$. Prove that$24 \leq |H \cup K| \leq 31$Since$H\ne K$, then$1\le\lvert H\cap K\rvert \le8\implies 24\le\lvert H\cup K\rvert \le31$. (BTW,$H\cup K$won't be a group.) 1 vote ### Finite extensions of finitely generated free groups. If$\Gamma$is your example$\mathbb Z_2\rtimes_{\varphi(\bar 1)} F_2$you are trying to extend$F_2$by$\mathbb Z_2$which means that you are trying to complete a short exact sequence $$0\to F_2\to\... 1 vote ### Do profinite completions commute with direct products? Yes, profinite completion commutes with finite products. Rather than wondering what a subgroup of the product group looks like, it's easier to verify the universal property directly. The basic idea is ... 1 vote Accepted ### How to Find Orbits and Stabilizers The formula you are using is incorrect. It is not used for the whole set, but for one element only$$\forall\ s\in S,\lvert D_6\rvert=\lvert\text{Stab}(s)\rvert\cdot\lvert\text{Orb}(s)\rvert$\$ That's ...

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