# Tag Info

## New answers tagged symmetric-groups

0

If you know that $A_5$ is simple and that all Sylow subgroups are conjugated, then this question is simple: since the set of all Sylow subgroups is invariant under conjugation, they generate a normal subgroup. So $K_{2,5}=A_5$ has order $60$. In fact, this will work more generally for finite simple groups and any prime $p$ dividing their order, for instance ...

4

We are looking for elements which commute with $a$, in other words we try to determine the centralizer $C_G(a)$, where $G=\mathbf S_8$. Here is an outline of what you can do: Step 1: the following elements will certainly commute with $a$: The cyclic group $\mathbf C_3$ generated by $(1\ 4\ 7)$; the involution $x=(2\ 5)$ the involution $s = (2\ 3)(5\ 8)$ ...

0

b commutes with a iff it is composed of similar disjoint cycles as a, except possibly up to singleton loops in either. In your example a has only a single such loop (6) so the only possible b are the 8 possible choices of a subset of the 3 nontrivial loops of a.

1

The fact that the action is transitive implies that $n$ divides the order of $G$. Cauchy's theorem then implies that $G$ contains an element of order $n$ because $n$ is prime. What are the elements of order $n$ in $S_n$?

2

Hint : how many transpositions have you in $S_5$? how many double transpositions have you in $S_5$?

3

It's not clear by your armgument, why there must exists a $\sigma$ for each $m$ such that $\sigma(m)=k$. On the other hand, there's no need for a counting argument. Given, $1\le m<k\le n$, let $\sigma$ be the transposition $(mk)$. That is, the bijection which switches $m$ and $k$ and leaves every other number fixed.

2

They do, they just don't give any symmetry groups we didn't already get from a Platonic Solid. For example, symmetric truncation preserves the symmetries -- so the truncated cube, truncated octahedron, and cuboctahedron all have rotational symmetry group $S_4$. In general, you'd just need to convince yourself that whatever operation we perform on a ...

0

For me the key point here is Cayley's theorem which tells us that every four element group is isomorphic to a subgroup of the symmetric group on four elements. Furthermore the proof tell us how to find which subgroups. The two four element groups are $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ choosing bijections $\phi$ and $\psi$ between their ...

2

The image of $\phi$ is a transitive subgroup of $S_{|G/H|}$, but that's all you can say about it; any transitive subgroup can appear (exercise). The kernel of $\phi$ is the intersection $\bigcap_{g \in G} gHg^{-1}$ of all of the conjugates of $H$ (exercise).

0

Note that $S_3$ is generated by $(1\ 2)$ and $(1\ 2\ 3)$: powers of either $(1\ 2)$ or $(1\ 2\ 3)$ account for $4$ elements of $S_3$ directly, and: $(1\ 2\ 3)(1\ 2) = (1\ 3)$ $(1\ 2)(1\ 2\ 3) = (2\ 3)$ account for the remaining $2$. So any homomorphism $\phi:S_3 \to S_3$ is completely determined by $\phi((1\ 2\ 3))$ and $\phi((1\ 2))$. Now an ...

1

For the future, you should work on growing a catalog of groups with which you're familiar. The two that come in handy here are The cyclic group of order $4$. If you can find an element $g$ of order $4$ in any group, then the subgroup $\langle g \rangle$ generated by $g$ is cyclic of order $4$. The other group that we'll care about is The Klein Four Group. ...

1

Consider $G_1=<(1 \ 2 \ 3 \ 4)>, G_2=<(1 \ 2),(3 \ 4)>$ $G_1$, $G_2$ are both of order 4 and are both subgroups of $S_4$, but $G_1$ is cyclic and $G_2$ isn't, hence they are non-isomorphic

0

If $d|24$, $d\in {1,2,3,4,6,8,12,24}$. $d=1: \{e\}$ $d=2:\{e,(1 2)\}$ $d=3:<(1 2 3)>$ $d=4:<(1 2 3 4)>$ $d=6: S_3$ $d=8:<(1 2), (2 3), (3 4)>$ $d=12: A_4$ $d=24: S_4$

2

The approach is right. For the non-cyclic subgroup, try the group generated by $(12)$ and $(34)$. It should be easy to produce a cyclic subgroup of order $4$.

2

For arbitrary subgroup $H$ of $G$, the homomorphism $\phi\colon G\rightarrow S_X=S_{G/H}$ is not necessarily injective or surjective. Consider $G$ with $|G|=9$ and $|H|=3$. We will get a homomorphism $\phi\colon G \rightarrow S_{G/H}\cong S_3$. Comparing orders, you will see that it is neither injective nor surjective.

2

One convenient way could be to consider only composite numbers: $2.3, 2^2.3, 2^3.3$. For divisor $2.3$, natural subgroup is $S_3$ (without looking the list, we can say, it is a natural candidate for subgroup of this order). For divisor $2^2.3$ again, natural subgroup $A_4$.

0

By definition $S_3=\{f:\{1,2,3\}\to\{1,2,3\}\,|\, f \text{ is invertible}\}$. Let $\sigma:\{1,2,3\}\to\{1,2,3\}$ be the map defined by $\sigma(1)=2$, $\sigma(2)=3$, and $\sigma(3)=1$, and let $\tau:\{1,2,3\}\to\{1,2,3\}$ be map defined by $\tau(1)=2$, $\tau(2)=1$, $\tau(3)=3$. A straightfoward calculation shows that $|S_3|=6$, and hence if we find $6$ ...

1

Think about the cyclic subgroup generated by the product of a $p$-cycle, a $q$-cycle, and a $r$-cycle, each disjoint with one another. What would the order of such an element, and the subgroup generated by it, be equal to?

4

Hint: $S_p$ contains a subgroup of order $p$. $S_p\times S_q$ is a subgroup of $S_{p+q}$. Note: As @pjs36 mentioned, there's no need for $p,q,r$ being prime. The extra you got from that hypothesis is the subgroup you find is cyclic.

2

Let me try to spell out what has been said in the comments. Showing that two groups are isomorphic involves exhibiting an isomorphism between them. As noted in the comments, you need to identify the groups involved: Let $X$ be a set. The group $S_X$ is the set of all bijections $X\to X$ with multiplication given by composition of functions. Now, let $X$ ...

4

Note that $\sigma$ has order 6 since the lcm of its cycle lengths in its cycle decomposition is 6. (Alternatively just check). Then $\langle \sigma\rangle \cong Z_6\cong \langle\tau\rangle$, so you just need to check whether or not they have trivial intersection. One method is to just write them out. But alternatively, note that $\sigma$ fixes 6, so that all ...

1

Lagrange's theorem states that the order of the subgroup must divide the order of the group. We know that $|S_n|=n!$, so in our case does the order of the subgroup divide $4!$?

4

If you mean, "is every subgroup of a symmetric group isomorphic to a symmetric group?", the answer is no. For example, look at the subgroup generated by a $k$-cycle, with $k>2$ inside of a symmetric group of order greater than $2$. This will be a cyclic group, and no symmetric group $S_n$ is cyclic for $n>2$.

1

Since $\pi^{20} = \pi^2$: $$\pi^2 = (17)^2(395)^2(486)^2 = (359)(468)$$

4

See http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf, all values of $n$ except 2, 5, 6, 8. The result is old, going back to Miller in 1901.

2

If $p |n$ and $p< n$ we have $$\frac{1}{n} \binom{n}{p} = \frac{(n-1)\dots(n-p+1)}{p!}$$ This cannot be an integer as $p| p!$ and $p \nmid (n-1)\dots(n-p+1)$.

0

Let's assume $n>1$ or the statement is trivial. It's well known that $\dbinom{n}{r}$ is divisible by $n$ for all $r$ with $0<r<n$, when $n$ is prime. Indeed $$\binom{n}{r}=\frac{n(n-1)\dots(n-r+1)}{r!}$$ is an integer and the denominator is not divisible by $n$, while the numerator is. Examining a table of the binomial coefficients makes one ...

1

There are two conventions you need to pick, and both of them affect the answer. The first convention is whether you read compositions in the standard order (which is right-to-left, in the sense that the rightmost permutation "happens first") or the other order. The second convention is whether permutations act on "numbers" or on "positions." For example, ...

0

Conceptually it is probably simplest to start by computing the two-line representation (last part of the equality you display), by tracing each element from right to left through the product of transposition. Once you have that, tracing the cycle structure of the permutation should be routine.

1

Note $s_1s_2s_1=(13)$. Thus we have $$(56)(45)(34)(13)=(65)(54)(43)(31)=(65431)=(16543)$$

3

This has nothing to do with odd permutations. Instead, you may consider the quotient map $p : S_n\rightarrow S_n/A_n = C_2$. Now ask yourself, what is the image of $G$ under this surjection? If $G$ surjects onto $C_2$, then it must have even order, thus $p(G) = 1$ in $C_2$, implying that $G$ is contained in the kernel of $p$, hence $G\le A_n$.

7

Suppose $G\nleq A_n, \exists g_0 \in G\setminus A_n$. Now consider $$A = \{g \in G : \text{sgn}(g) = 1\}, \text{ and } B = \{g \in G : \text{sgn}(g) = -1\}$$ Then $A = g_0B$, so $|A| = |B|$. Also, $G = A\sqcup B$, so $|G|$ would be even. This is a contradiction.

4

12=lcm(2,3,4), and the perm has cycle structure 10=1+2+3+4, is even. 10 = lcm(2,2,5) and the perm has cycle structure 10= 1+2+2+5 is even. To get an element of order 2, e.g., take two disjoint transpositions Others can be done similarly.

1

I'm not sure why you would want to use unlabeled species here. It appears that the standard procedure is to observe that the OGF of the cycle index of the symmetric group is $$G(z) = \sum_{q\ge 0} Z(S_q) z^q = \exp\left(a_1 z + a_2 \frac{z^2}{2} + a_3 \frac{z^3}{3} + \cdots\right).$$ We are interested in the coefficient $$n! [z^n a_1^{k_1} a_2^{k_2} ... 1 You just have to use the formula for the conjugate of a cycle by a permutation \sigma:$$\sigma(1\,2\,4\,3)\sigma^{-1}=\bigl(\sigma(1)\,\sigma(2)\,\sigma(4)\,\sigma(3)\bigr)=(2\,3\,1\,4)=(1\,4\,2\,3).$$Justification:$$\sigma(1)\mapsto \sigma^{-1}\sigma(1)=1\mapsto2\mapsto\sigma(2).$$0 Each element of GL(2,\mathbb F_2) can be viewed as a linear transformation f on \mathbb F_2^2, which sends$$(1,0)\mapsto x, (0,1)\mapsto y.$$It follows that f(1,1)=x+y is the only other non-zero vector. Thus f defines a permutation \sigma_f of the three non-zero vectors of \mathbb F_2^2. The mapping$$f\mapsto \sigma_f$$is clearly a group ... 2 The vector space \mathbb{F}_2^2 has exactly three nonzero elements. So, given some ordering of these, there is a clear group homomorphism GL(2,\mathbb{F}_2)\to S_3. Since the two groups have the same number of elements, it is enough to show that this map is injective. 1 Your proof would be something like Let s_i be in \{s_1,s_2,\ldots,s_n\} and t_j be in \{t_1,t_2,\ldots,t_m\}. Then \sigma\circ\tau(s_i)=\sigma(\tau(s_i))=\sigma(s_i)=s_{i+1} and \tau(\sigma(s_i))=\tau(s_{i+1})=s_{i+1}. Similarly for t_j. Also note both functions have the same domain. Thus the two functions are the same. 0 For each n-cycle w, it has a representation (a_1a_2...a_n). w^{n-1}(a_1)=a_{n}\neq a_1 so w has order at least n. 1 If we let D_n act on the set of n vertices of a regular n-gon, this gives us a homomorphism: D_n \to S_n. This action is faithful, as the only symmetry of the n-gon which fixes all the vertices is the identity map. This shows D_n \subseteq S_n. However, |D_n| = 2n < n! = |S_n| if n > 3. 0 If you compose "right-to-left" this is pretty easy to see: x_1 is unmoved by all transpositions except the left-most, so that the net result is x_1 \mapsto x_2. Similarly, x_2 is unmoved by all but the next-to-left-most, where it it gets sent to x_3, which is unmoved by the left-most transposition (x_1\ x_2), so that the net result is x_2 ... 1 HINT: Prove it by induction on r. The key step is to show that$$(x_1,\ldots,x_{r-1})\circ(x_{r-1},x_r)=(x_1,\ldots,x_r)\;, which is pretty straightforward.

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