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## New answers tagged group-theory

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### Criterion for cyclic groups in terms of its number of subgroups

I found a proof (from an old sci.math post of my own) that a finite group of order $n$ has at least $d(n)$ (the number of divisors of $n$) subgroups. This follows from the case $m=n$ in the result ...
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### uniqueness of generators of Lie groups

Qiaochu has already answered this in the comments but I wanted to put together a longer answer as I see this confusion fairly regularly here. It is quite common in Physics to use "the" a bit ...
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• 93.2k
3 votes
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(All vector spaces, algebras, etc. below are real and finite-dimensional.) So, let's discuss the following general question: suppose $A$ is a vector space equipped with a multiplication $m : A \otimes ... • 431k 6 votes Accepted ### Is a transitive-on-unit-vectors subgroup of$SO(3)$automatically all of$SO(3)$? First of all, the group$G=SO(3)$is simple as an abstract group: This fact was discussed several times on this site (see, say, here). Let's prove that your group$F$is a normal subgroup of$G$, i.e. ... • 101k 9 votes ### Is there a general way to find the inverse of an automorphism of the free group? Yes. Let$X$be the given (ordered) free generating set of the free group, and let$Y$be the set of images of the elements of$X$under the automorphism. Now perform Nielsen reduction on$Y$, and ... • 92.1k 4 votes Accepted ### A group with exactly half of the elements in one conjugacy class Your construction in paragraph 2 gives all examples! Now the conjugacy action$G$on$S$gives an injective map$i:G\to S_{|G|/2}$. I follow you up to here but here I think you've made a small extra ... • 431k 1 vote Accepted ### Derived series of a square-free order group stabilizes Since${\rm Aut}(G''/G'')$is abelian, the restriction to$G'$of the action of$G$on$G''/G'''$is trivial, so$G''/G''' \le Z(G'/G''')$. But$G'/G''$is a direct product of cyclic groups of prime ... • 92.1k 3 votes ### How do you prove associativity for this operation? Short answer $$\frac{a/(1/b)}{1/c}=\frac{(a/(1/b))\color{blue}{/(1/(1/b))}}{(1/c)\color{blue}{/(1/(1/b))}}=\frac{(a\color{red}{/(1/b)})/(1\color{red}{/(1/b)})}{(1/c)/(\color{red}{1/(1/b)})}=\frac{a}{(... • 7,116 0 votes ### S_6 contains two subgroups that are isomorphic to S_5 but are not conjugate to each other Transitivity of a subgroup will be preserved by conjugation. There's 6 obvious non-transitive S_5's in S_6. There's also famously a transitive S_5, corresponding to an outer automorphism. ... • 9,388 4 votes Accepted ### Possible indices of finite index subgroups of SL_2(\mathbb{Z}) Actually we can give a straightforward uniform argument that \Gamma \cong PSL_2(\mathbb{Z}), and hence SL_2(\mathbb{Z}), has a subgroup of index n for every positive integer n. The sequence of ... • 431k 2 votes Accepted ### Relative divisibility of derived subgroup of free group No. If F has rank \kappa (any cardinal, possibly infinite), then quotient F/[F,F] is the free abelian group of rank \kappa. As free abelian groups are torsion free, if w^n\in [F,F], then w[... • 404k 3 votes Accepted ### Show that the given representation of the group G is reducible Whenever you feel stuck on a question, a basic reflex you need to have is to write down the definitions and see if you understand them. Very often in mathematics, questions are answered effectively by ... 2 votes Accepted ### Let D_6 = \langle a, b \mid a^6 = b^2 = e, ba = a^5b \rangle be the dihedral group Your solution for part (a) is correct. Part (b): You have been asked to find all subgroups of order 4 in D_6. Your table shows that there is no element of order 4 in D_6. So, we do not have ... • 2,262 0 votes ### Let D_6 = \langle a, b \mid a^6 = b^2 = e, ba = a^5b \rangle be the dihedral group Your answer for (a) is correct. Your answer for (b) is not correct. For example number 2 fails because b \cdot ab = a^5 which is not an element of the subgroup. First to get a subgroup S of order ... • 1,568 5 votes ### A_{11} has no subgroups of order \frac{11!}{14}? SOLUTION: Assume by way of contradiction that there exists H \leq A_{11} with |H| = 11!/14. Let A_{11}/H be the set of left cosets of H in A_{11}; e.g. the set of all \sigma H for \sigma \... 2 votes Accepted ### Let a be the reflection of the plane \mathbb{R}^2 over the bisector of the odd quadrants For part (a), note that G = \langle a, b \rangle indicates the subgroup generated by a and b (in which larger group?). So, in general, this can be rather large, since any word we write down in ... • 26.1k 3 votes ### Understanding proof of existence of Schreier transversals Your question is answered by a standard application of Van Kampen's Theorem, and the set J is defined in the second sentence of the second paragraph of the proof. Perhaps it will be clearer if one ... • 124k 12 votes Accepted ### Group of order n is a subgroup of S_{n-1} Yes your idea works exactly as stated. Since n is not a prime power it has distinct prime divisors p and q and subgroups H and K. Let \phi_H and \phi_K be the homomorphisms of G into ... • 92.1k 3 votes Accepted ### Describe all non-isomorphic groups of order 57 As noted in the comments, |G| = pq a product of distinct primes doesn't imply G is cyclic (for example, S_3 has order 2 \cdot 3), so the attempt in the question fails. If |G| = 3 \cdot 19, ... • 7,799 2 votes Accepted ### What is the index (G : C_G(B)) ? The characteristic polynomial of B is (x-1)^2, so it admits a Jordan decomposition over the field with 5 elements, with Jordan normal form J=\begin{pmatrix}1&1\\0&1\end{pmatrix} (... • 36k 4 votes Accepted ### The Uniqueness of the Logarithm as a Group Isomorphism between the Positive Reals and Reals This is false without more hypotheses on \varphi, e.g. it suffices to require that \varphi is continuous, or monotonic, or measurable. Without any hypotheses we can compose \varphi with ... • 431k 0 votes ### "Abstract" presentation of SL(2,\mathbb Z) The group SL(2,\mathbb Z) does not have a presentation of the form that you wrote. If you substitute U=ST into that presentation, then you can eliminate T and rewrite your presentation in the ... • 124k 2 votes ### A certain inverse limit Yes. It is a consequence of Galois theory : Let L|K be a finite Galois extension, of Galois group G. Then there is a decreasing bijection between distinguished subgroups of G, and subextensions ... 5 votes Accepted ### Do sets of commuting permutations with no fixed points generate Abelian groups with no fixed points? It is difficult for permutation actions to have the property that no non-identity element has fixed points. This property is called being free, and for a group G acting on a set X it is equivalent ... • 431k 3 votes Accepted ### subgroups of (\mathbf Q, +) as direct limits Yes, that's right. The general pattern is that the directed colimit of$$\mathbb{Z} \xrightarrow{n_1} \mathbb{Z} \xrightarrow{n_2} \mathbb{Z} \xrightarrow{n_3} \dots $$computes an increasing union of ... • 431k 0 votes ### Show that no group of order 48 is simple Suppose that G of order 48 is simple and n_2=3. Therefore, G\hookrightarrow S_3 (consider some G-action on the quotient set G/P_2, where P_2 is any Sylow 2-subgroup): contradiction, by ... • 3,393 1 vote ### How to prove that all elements inside a cycle of a cyclic group are different from each other Here is the standard way to see this. Suppose that n = ord(a), so n is the smallest positive integer such that a^n = e. Then note the following: (*) a^k = e if and only if n divides k. ... • 1,033 3 votes Accepted ### Irreps of SU(3)/\mathbb{Z}_3 from irreps of SU(3) (All representations are complex and finite-dimensional throughout except for at a handful of points where I talk about real representations.) In general, if V is an irreducible representation of a ... • 431k 6 votes ### Surjective homomorphism \mathbb Z * \mathbb Z \to C_2 * C_3; can my proof be rescued? The key is that the coproduct is generated by the canonical images of the groups it is a coproduct of. Lemma. Let G_1 and G_2 be groups, and let \iota_j\colon G_j\to G_1*G_2 be the canonical ... • 404k 1 vote Accepted ### Groups of homeomorphisms vs Configuration spaces First of all, for every connected manifold M (without boundary), Homeo(M) acts transitively on the n-fold configuration space of M: This was discussed several times on MSE. It follows that ... • 101k 9 votes Accepted ### Does there exist a group G such that \operatorname{Aut}(G)\cong D_5, where D_5 denotes the dihedral group of order 10? There is no such group. First, as already laid out in the comments, G cannot be abelian. For completeness sake, here is the argument again: If G is abelian and not an \mathbb{F}_2-vector space, ... • 2,243 0 votes ### Show that no group of order 48 is simple Analyzing the 2-Sylows: The number n_2 of 2-Sylows is of the form n_2 = 2k + 1, for some integer k \geq 0, where n_2 \mid 3. Testing the possibilities, we see that (n_2, k) = (1, 0), (3, 1). ... 2 votes Accepted ### Let \mathbb{R^*} be the multiplicative group of nonzero real numbers.Which of the following statements are true? For statement 2, if x^3 \in H for all x \in \mathbb{R}^*, then x=(\sqrt[3]{x})^3 \in \mathbb{R}^* for all x \in \mathbb{R}^* (using the existence of cube roots). For statement 4, suppose ... • 10.1k 4 votes Accepted ### Is this definition of a cycle in symmetric groupsâ€‹ correct?" You've correctly identified a very common (and useful) abuse of notation. When a small cycle is written in a "large" symmetric group, the implicit assumption is that all numbers not ... • 24.3k 3 votes ### Is this definition of a cycle in symmetric groupsâ€‹ correct?" Yes it is correct. The cycle (2 4) tells you that something is happening to the numbers 2 and 4. Implicitly, all other numbers do not change, so the cycle bijection keeps them fixed, and it is truly a ... 1 vote Accepted ### Why is the order of an element equal to the order of the group it generates? Suppose that ord(a) = n for some integer n > 0. Then a^n =e. Thus a \circ a^{n-1} = e, and multiplying both sides with a^{-1} gives you$$a^{-1} = a^{n-1}.$$Now raising both powers by ... • 1,033 1 vote Accepted ### Show that H is a normal subgroup of G. Consider the action of G in G/H (note that it isn't necessary a group) by translation, x.(gH)=(xg)H. We know \#G/H=3, then we have a morphism from G to Biy(G/H)\cong S_3 by x\mapsto \... 1 vote ### Is every (infinite) permutation the composition of 2 involutions in ZF? I don't have an answer, but I want to note we know a lot about the structure of the involutions. Let \tau(x)=y. Then we have$$\sigma(y)=f(x)\sigma(f(x))=y$$so$$\tau(f^{-1}(y))=f(x)\tau(... • 5,488 1 vote ### How to find the order of$\text{Aut}(\text{Aut}(\mathbb{Z}_{1080}))$This should be a comment but I need the nice formatting of an answer, so I have made it CW. According to GAP: ... 4 votes Accepted ### How to find the order of$\text{Aut}(\text{Aut}(\mathbb{Z}_{1080}))$No, in general the group${\rm Aut}({\rm Aut}(\Bbb Z/n))$does not have order$\phi(\phi(n))$. The group need not even be abelian. It is useful to look at a smaller example first. Consider$n=12$. ... • 133k 1 vote ### Can the sum of a nonlinear irreducible character's values on$Z(\chi)$be zero? This sum of values is not equal to zero if and only if$\chi$is constant over$Z(\chi)$. First, write the restriction of$\chi$to$Z(\chi)$as$\chi_{Z(\chi)}=\chi(1)\lambda$, where$\lambda$is a ... • 1,176 4 votes Accepted ### If$G/Z(G)$is isomorphic to a subgroup of$\mathbb Q$then$G$is abelian. Hint: assume for contradiction that some$a, b \in G$don't commute, and consider$H = \left< a, b \right> \leqslant G$. PS It is also not hard to unpack the reasoning so that it does not use ... • 11.1k 3 votes Accepted ### Let$G$be a group of order$p^nq$where$p$and$q$are distinct primes and suppose$q \nmid p^i-1$for$1 \leq i \leq n-1$. Prove$G$is solvable I'm afraid almost everything is incorrect You are correct that the number of$q$-Sylow subgroups must be$1$. However, your argument via the claim "but$\mathbb{Z}_q$is abelian, hence normal, ... • 404k 2 votes ### Inclusions of product of groups Let me denote the cyclic group of order$k$by$C_k$, written multiplicatively. Claim. Let$G_2=C_n\times C_n$, with the factors generated by$x$and$y$, and let$m$be a positive divisor of$n$. The ... • 404k 2 votes ### Prove the binary icosahedral group is isomorphic to${\rm SL}(2,5)$Your strategy looks fine to me, let's try to make it work. First, working$\bmod \sqrt{5}$we can invert$2$and$\tau \equiv 3 \bmod \sqrt{5}$, so we get$\zeta \equiv -1 + i + 3j \bmod \sqrt{5}$. So ... • 431k 3 votes Accepted ### Conjugacy classes of normal subgroup in group You need every element of$N$to have centralizer not contained in$N$. There are examples in which$C_G(N)$is contained in$N$. A computer search shows that the smallest such example has order$64\$ ...
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