# Tag Info

6

In general, if the prime $p$ divides $n$, then the permutation representation of $S_n$ over the field ${\mathbb F}_p$ has two irreducible constituents of degree $1$ and one of degree $n-2$ and, for $n \ge 5$, this results in an embedding $S_n \to {\rm GL}(n-2,p)$. Also, $S_4$ embeds into $GL(2,{\mathbb Z}_4)$: you can take the images of $(1,2)$ and ...

4

By the standard theorems on covering spaces, conjugacy classes of maps $\varphi : F_n \to S_k$ are naturally in bijection with $k$-fold covers $Y$ (not necessarily connected) of a wedge $X = \bigvee_{i=1}^n S^1$ of $n$ circles. The cycle type of the image of a generator of $F_n$ tells you what the preimage of the corresponding copy of $S^1$ should be in this ...

4

Your question (and the answer you gave) show that you've missed an essential point. In group theory at a very early point one remarks that the associativity axiom can be iterated to show that any two pure product expressions (no inverses) of the same string of variables, but grouped differently by parentheses to show which sequence of multiplications is ...

4

The additional generator you could take to generate a Sylow $p$-subgroup of $S_{p^2}$ is $(1,p+1,2p+1,\ldots,(p-1)p+1)(2,p+2,2p+2,\ldots,(p-1)p+2) \cdots (p,2p,3p,\ldots p^2)$ For example, for $p=3$, this would be $(1,4,7)(2,5,8)(3,6,9)$. This normalizes the elementary abelian subgroup of order $p^p$ that you have already found: check that it conjugates ...

4

I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way: Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups. Similarly, the Sylow-3 subgroups ...

4

Think about a circular conveyor belt where $n$ boxes are running in circles constantly. Some guy is standing at a fixed position at the belt and can only swap the two boxes right in front of him. By waiting for different boxes to come by he can swap any two neighboring boxes. Can he arrange the $n$ boxes in any given order? How does this relate to ...

3

As the following shows, GAP can work out generators of the kernel. Perhaps you were put off by the fact that it said "no generators known". That just means that it has not calculated them yet. The other possibility is that the example you gave it was too large. If you tried mapping $F_{20}$ to $S_{20}$ for example, then $K$ would have far too many ...

2

To find Sylow $p$-subgroups in a group in which you can generally find $p$-elements and normalizers the following algorithm is sometimes used: Assume $P_i$ is a $p$-subgroup of $G$ (we can take $P_0 = \{1_G\}$ to be the trivial subgroup). If $P_i$ is a Sylow $p$-subgroup, then yay, return $P=P_i$. Otherwise we know by theory* that $P_i$ is not a Sylow ...

2

Just a sketch of a proof; sorry, this is all I can do with the time I have right now. If $m\in\mathbb N$ is arbitrary, then an $m$-reciprocal polynomial means a polynomial $p \in \mathbb Z\left[q\right]$ whose coefficient before $q^i$ equals its coefficient before $q^{m-i}$ for every $i\in\mathbb Z$ (this implies that the degree of $p$ is $\leq m$). Let ...

2

Let $f:C_8\rightarrow S_3$ be non-trivial homomorphism. Then one can see that $\frac{C_8}{Kerf}$ is isomorphic to a subgroup of $S_3$. Note that $|\frac{C_8}{Kerf}|$ is power of $2$. Thus, $\frac{C_8}{Kerf}\cong \langle \sigma \rangle$ where $o(\sigma)=2$. Now, Assume that $C_8=\langle x \rangle$. We can define homomorphisms $f_1(x)=(1,2)$, $f_2(x)=(1,3)$ ...

2

You ask about the element $(2,3,1)$. What this element means is that the second position is sent to the third position, the third to the first, and the first to the second. But notice this is the same as $(1,2,3)$ and $(3,1,2)$. So $(2,3,1)$ is in $S_3$, and you'd already mentioned it - but you called it $(1,2,3)$ above.

2

I fill in Ted's exquisite answer. This is tricky - the two cases look the same, but they're not. The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$, while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. To see why, consider these 2 permutations: $\color{tomato}{p_1(1) = 1, p_1(2) = 3, ... 1 It seems to be easier just to observe that the six sides of one of the hexagons are not all equal -- three of them border on pentagons and three of them border on hexagons. So a 60° rotation like you describe would take pentagons to hexagons and vice versa -- it wouldn't be a symmetry at all. 1 In a word: symmetry. For example, you may have already encountered the set of all symmetries of the square: all the ways of placing a square on top of itself. There are 8: four rotations (including the 360 degree rotation), plus four reflections (flips). These form a group under composition (first do one symmetry, follow it with another). The frieze groups ... 1 I won't calculate the exact result but let me give you some ingredients: You should know how to transform a matrix to another via a change of basis. This way we can work with a preferred one. Mine is the standard basis. Since you used$e_i$already lets call them$v_1,v_2,v_3$. You hopefully know how the permutation matrices that you called ... 1 Let$\sigma_1,\sigma_2,...,\sigma_p$be a collection of pairwise commuting$p$cycles in$S_{p^2}$, where$\sigma_i =(a_{i,1},a_{i,2},..,a_{1,p})$. Choose a$p^2$cycle$\tau =(a_{1,1},a_{2,1},..,a_{p,1},a_{1,2},a_{2,2},..,a_{p,2},...,a_{1,p},a_{2,p},..,a_{p,p})$. Clearly this$\tau^p=\sigma_1.\sigma_2..\sigma_p \in <\sigma_1,\sigma_2,...,\sigma_p>\$ ...

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