New answers tagged

2

There is indeed no such homomorphism. An easy way to check this is to write down the size-$2$ subgroups of $S_4$, take the quotient of $S_4$ by them, and then observe that you never get $A_4$. Alternatively, it suffices to prove that no size-$2$ subgroup of $S_4$ is normal. This is obvious because any normal subgroup containing a transposition must contain ...


3

Just because you can view a group as a subgroup of another (i.e., via an isomorphism with the subgroup) doesn't mean that it really is a subgroup. For a perhaps more enlightening example, consider $\mathbb{Z}/2$. This is not a subset of $\mathbb{Z}/4$ in the strict sense of the term, but if we see the former as $\{0,1\}$ and the latter as $\{0,1,2,3\}$, it ...


0

On your second bullet: If the intersection is not trivial then it contains an element that has order $p$. This elements generates both subgroups so they will coincide. Derek was dealing however with distinct subgroups.


0

For your first question: Let $\sigma \in S_p$. Then $\sigma$ can uniquely be written as a product of disjoint cycles. The concatenation of the lengths of these cycles is called the cycle-type of $\sigma$, for example, the cycle type of $(12)(34)$ in $S_5$ is $2$-$2$-$1$. Let $m_1$-..-$m_k$ be the cycle type of $\sigma$. Then $\text{order}(\sigma)=\text{lcm} (...


0

First point: a cycle in $S_p$ is usually denoted with the smallest element encountered in the cycle as firs element. As it is a cycle of length $p$, the smallest element is $1$. For the second point, in any group, two subgroups of prime orders intersect trivially because of Lagrange's theorem: the order of the intersection is the g.c.d. of the orders, so ...


0

A very nice proof is the following, due to Galois. Let $N$ be a normal subgroup, and suppose that it is nontrivial. Pick an element $\sigma$ with the maximum number of fixed points that is not the identity. If it is not a three cycle, its cycle decomposition is of the form $\sigma =(123\cdots)\cdots$ or $\sigma =(12)(34)\cdots$. If you think about it for a ...


0

Here is a proof of the first assertion. Let $a_i$ be an increasing or decreasing sequence. For a permutation $\sigma\in S_n$, consider the sum $$S(\sigma)=\sum_{i=1}^na_ia_{\sigma(i)}.$$ An inversion is a pair $(j,k)\subset\{1,\dots,n\}$ with $j<k$ and $\sigma(j)>\sigma(k)$. All non-identity permutations have at least one inversion. Take the minimum ...


0

I recommend you "A COURSE ON GROUP THEORY" by John S. Rose. You can find it here. Unfortunately, it is not well written, however it contains plenty of results. Moreover, it is not introductory and it focuses on the notion of group actions on sets and on groups, see Burnside Theorem and else. (I don' t know if you can find it free in online form)


2

I think one of the best and most complete book I know on (especially finite) group theory is Antonio Machì's http://www.springer.com/us/book/9788847024205 (original language: italian), despite the word "Introduction" in the subtitle. The author is also the translator of Herstein's much more elementary $\textit{Algebra}$. There are also many exercises and ...


-1

Suppose you have a cycle $C=(c_1,c_2\dots c_n)$ of $\sigma$. then after conjugating it you must get another cycle of the same size. Of course, since $\sigma$ is equal to its conjugate and all of its cycles have different lengths, we conclude $C$ is equal to its conjugate. And therefore, by your lemma we have $\tau^{-1}(c_1),\tau^{-1}(c_2)\dots \tau^{-1}(c_n)...


1

Two permutations have the same cyclic structure (i.e. they decompose into the product of the same number disjoint cycles of the same length) if and only if they are conjugated. So for any $p,q\in S_n$ you have $qp=q(pq)q^{-1}$ and hence they are conjugated. So no, it is not a coincidence.


0

Please take a look at this tutorial posted at this link. http://blogs.mathworks.com/graphics/2016/01/29/tiling-hexagons-and-other-permutohedra/ I think it does exactly what you want but incorporates graph theoretical concepts for a cleaner presentation.


3

From a purely pedantic formal perspective: No, merely defining the bijection between the permutation set and A does not define a group; you need to also provide the group operation as a relation on A. Of course, I understand that what you intend is implicitly obvious, but implicitness doesn't count in definitions. The point is that, formally, you don't ...


3

If I understand you correctly, the relation $R$ you describe is not just like a bijection, it is a bijection. Specifically, $(A,R,B)$ with $R\subseteq A \times B$ is a function if: $$\forall a\in A : \exists! b\in B : (a,b)\in R$$ and such a function is called a bijection, if: $$\forall b\in B : \exists! a\in A : (a,b)\in R$$ It is not surprising, that $A$ ...


2

There are infinitely many positive integers $n$ such that $S_n$ has an abelian subgroup of order $n^2-1$. Suppose $n$ is odd. Let $u\in\{-1,+1\}$ be such that $n\equiv u\pmod{4}$. Then, $$n^2-1=2^{k+1}\,\left(\frac{n-u}{2^k}\right)\,\left(\frac{n+u}{2}\right)\,,$$ where $2^k$ is the largest power of $2$ that divides $n-u$ (whence $k\geq 2$). If $$2^{k+...


4

It can't exist in general, in fact if $n=p-1$ where p is prime, then there does not even exist a sub-group of order $n^2-1$ because the order of a subgroup must divide that of the group and $\frac{(p-1)!}{(p-1)^2-1} = \frac{(p-1)!}{p(p-2)}$ is not an integer.


1

Since by definition $D_n$ has an element of order $2$, $D_n$ is a $p$-group iff it is a $2$-group iff $2n$ is a power of $2$ iff $n$ is a power of $2$.


2

$S_n$ admits a subgroup (not necessarily symmetric) of order $n^2-1$ if and only if neither of $n\pm 1$ is prime. In particular, it admits such a subgroup if $n$ is odd and at least $5$.


2

This is only a partial attempt at tackling a more general underlying question. Note that if $n \ge 5$ is odd, then $n^{2} - 1$ divides $n!$. In fact $n^{2} = (n - 1) (n +1)$. Clearly $n-1$ divides $n!$. As to $n+1$, since $n$ is odd, $n+1$ is even, so $$ n+1 = 2 \cdot \frac{n+1}{2}. $$ Clearly also $2$ and $\frac{n+1}{2}$ divide $n!$, Now if $n \ge 5$, we ...


7

For $m\geq n$, the group $S_m$ contains a copy of $S_n$. For the given numbers if you calculate $n^2-1$ you get $4!,5!$ and $7!$ respectively. Thus they all have such subgroups. You can compute these subgroups very easily. For example $$S_4=\{1, (12), (13), (14), (23), (24), (34) (12)(34), (13)(24), (14)(23) (123), (124), (132), (134), (142), (143), (234), (...


2

Any $n \geq 2$. Partition according to parity so that $A = A_n$ and $B = S_n - A_n$. Take $\pi = (1, 2)$, which is an odd permutation. Since any even permutation in $A$ multiplied by the odd permutation $\pi$ will result in an odd permutation in $B$, we are done.


2

No: For any infinite set $X$, the cardinality of $S(X)$ is the same as that of its power set $\mathcal P(X)$ and therefore (by Cantor's theorem) different from the cardinality of $X$ itself. We need to know that there is a bijection $f: X\times X\to X$. That this exists for every infinite $X$ is a well-known consequence (and, indeed, equivalent) of the ...


1

The answer is $n = \sum_{k} p_k^{n_k}$. That follows from the more general result that, for finite nilpotent groups $G$ and $H$, ${\rm mindeg}(G \times H) = {\rm mindeg}(G) + {\rm mindeg}(H)$. I haven't checked the reference myself, but it is proved in D. Wright. Degrees of minimal embeddings of some direct products. Amer. J. Math., 97:897–903, 1975. See ...


2

Chapter 27 of Babai's "Handbook of Combinatorics" discusses automorphism groups, and in particular section 4 covers the universality problem. He does provide a number of references. You can access the chapter here. Trivalent graphs: R. Frucht, "Graphs of degree 3 with given abstract group", Canad. J. Math. 1 (1949), 365-378. Regular graphs (valency $k \...


1

Yes, your solution is correct. The number of ways to choose the three fixed points is ${8 \choose 3}$, the number of ways to choose the transposition from the remaining $5$ points is ${5 \choose 2}$, and the number of ways to form a $3$-cycle from the remaining $3$ points is $(3-1)!=2$. Thus, the number in question is ${8 \choose 3} \cdot {5 \choose 2} \...


1

In $|S_{20}|$, the highest power of $7$ which divides $20!$ is $7^2$. So it is clear that the Sylow-$7$ subgroup of $S_{20}$ is of order $7^2$. Group of order $7^2$ is either cyclic or isomorphic to $Z_7\times Z_7$. If it is cyclic, then $S_{20}$ will have an element of order $49$, and it should be product of disjoint cycles. Check, whether this is ...



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