Tag Info

2

The notation $|\mu|$ means the order of $\mu$ in $S_9$. Hint: Every permutation is a product of disjoint cycles The order of a product of disjoint cycles is the lcm of their orders Solution:

0

Seems like your problem is the exponentiation of permutations. I explain the first computation $\alpha^{121}$. First, do a integer division of the exponent by the order of $\alpha$: $$121 = 10\cdot 12 + 1$$ So $$\alpha^{121} = \alpha^{12\cdot 10 + 1}$$ Now by the exponentiation rules $$\alpha^{12\cdot 10 + 1} = (\alpha^{12})^{10} \cdot \alpha$$ Since the ...

1

Everything you've done, you've done correctly, as far as I can see. Here are a few comments: For 1 your computation of $\alpha^{121}$ is correct. An extra step to throw in if you're feeling uncertain might be $(\alpha^{12})^{10}\alpha^1= e^{10}\alpha=\alpha$, since $\alpha^{12}$ and $e$ are the same thing ($e$ is the identity element, the trivial ...

1

If $\varphi$ is an isomorphism $G \rightarrow S_n$, then the pre-image of $A_n$, i.e., $$H:=\varphi^{-1}(A_n)=\{g \in G:\varphi(g) \in A_n\}$$ is a subgroup of $G$ isomorphic to $A_n$ (the isomorphism is $\varphi$ with its domain restricted to $H$). This same technique works for any subgroup of $G$.

1

By looking at the possible cycle types, we see that $A_4$ consists of the identity element (order $1$), $3$ double transpositions (order $2$) and $8$ $3$-cycles (order $3$). Assume that $A_4$ has a subgroup $H$ of order $6$. Since $A_4$ does not contain elements of order $6$, $H$ cannot be cyclic. Therefore $H \cong S_3$, implying that $H$ contains $3$ ...

0

G= disjoint union of classes [x] under the conjugate relation. Any x aside from the identity element has only one of the forms: (12) with 6 elements , (123) with 8 elements , (1234) with 6 elements, (12)(34) with 3 elements elements. Thus the cardinalities of the nontrivial classes are 6,8,6,and 3. We are left with only one trivial class of one element and ...

0

Hint: The centraliser of $(12)(34)$ in $S_4$ is given by what you have shown: $$C_{S_4}((12)(34))=\{g\in G| g(12)(34)g^{-1}=(12)(34)\}$$ Firstly we note that by cycle type, conjugating it with anything will give you back a $2-2$ cycle type(since these are all of the things in the conjugacy class), so now we just need to find what doesn't shift it to one of ...

1

I am not sure if I am understanding your question correctly. I think it is the following. Let $G$ be a transitive subgroup of $S_n$ with $n$ even. Does there necessarily exist an element of order $2$ in $G$ that fixes at most $2$ points? I am afraid that the answer to that is no. In the terminology of the GAP and Magma computer algebra systems, the group ...

1

The number $$\binom{12}{2}\binom{10}{3}\binom{7}{5}$$ counts the ways to pick numbers that belong to each cycle. We must also specify how the numbers are arranged in the cycle. The number of ways to arrange $k$ numbers is $k!$, but cyclic permutation of an arrangement does not change the $k$-cycle, so since there are $k$ cyclic permutations of $k$ numbers ...

1

Rather than the action on the $3$-Sylow, I think there is a more natural set on which $SL_2(\mathbb{Z}_3)/Z(SL_2(\mathbb{Z}_3))$ acts faithfully. Set $X:=\{\text{ vectorial lines in } \mathbb{F}_3\times \mathbb{F}_3\}$. The cardinal of $X$ is easy to find, indeed, any vectorial line is given by a non-null vector and furthermore two colinear vectors give ...

2

This conjecture is false. Choose a multiple $d=k n$ of $n$, with $k>1$, such that $n(n-1)/2$ divides $d$ (that is, $d/\binom{n}{2} = 2k/(n-1)$ is an integer $r$.) The monomial $x_1^k\cdots x_n^k$, which has total degree $d$, can be divided up into $n(n-1)/2$ degree-$r$ factors in any number of arbitrary ways, having no invariance properties whatsoever. ...

1

This conjecture is false. Choose a multiple $d=k n$ of $n$, such that $n(n-1)/2$ divides $d$ (that is, $d/\binom{n}{2} = 2k/(n-1)$ is an integer $r$.) The monomial $x_1^k\cdots x_n^k$, which has total degree $d$, can be divided up into $n(n-1)/2$ degree-$r$ factors in any number of arbitrary ways, having no invariance properties whatsoever.

0

Coming late to this, but it's worth pointing out that there is an interesting proof by induction that avoids characters and whatnot, "from scratch", in Hernández-Lamoneda, L.; Juárez, R.; Sánchez-Sánchez, F. Dissection of solutions in cooperative game theory using representation techniques. Internat. J. Game Theory 35 (2007), no. 3, 395–426. Let ...

3

Let us formalize it the following way : $$Y=X_1\cup X_2$$ This is a disjoint union. Set $S(*)$ to be the symmetry of $*$. We know that $S(X_1)=S(X_2)=G$. Then $S(Y)$ acts on $Y$, set $S_1$ the set of symmetry fixing $X_1$ then they fix $X_2$, hence $S_1$ is $S(X_1)\times S(X_2)=G^2$. Now we have that $S(Y)/S_1$ is the group of permutation of ...

3

In fact in this case the order statistics do give you enough information to prove that the group is isomorphic to $S_4$ but I agree with Jack Yoon that this may not be the best approach. A group of order $24$ has $1$ or $4$ Sylow $3$-subgroups, and the fact that there are $8$ elements of order $3$ shows that there must be $4$. The image $P$ of the ...

1

After reading the first part, I was just about to suggest the method of your last paragraph. This is the preferred method to show the desired result, I suppose. For example, the symmetry group of a cube has $24$ elements because we can pick a face and for this face four different orientations. The fact that the group is isomorphic to $S_4$ becomes ...

0

The following solution only needs basic group theory. Let $G$ be an transitive abelian subgroup of $S_n$. By transitivity, for each $i\in\{1,\ldots,n\}$ there is a $\sigma\in G$ such that $\sigma(1) = i$. So $\# G\geq n$. Assume that $\#G > n$. Then there are $\sigma, \tau\in G$ with $x := \sigma(1) = \tau(1)$ and $\sigma\neq \tau$. By the second ...

0

Two cycles are said to be disjoint if the sets of elements they act on are disjoint. This is equivalent to saying that $(a_1\ \dotsc\ a_k)$ and $(b_1\ \dotsc\ b_h)$ are disjoint if and only if $a_i \neq b_j$ for every $1 \leq i \leq k$ and $1 \leq j \leq h$. What you misunderstood is cycle decomposition: $$(1\ 3\ 7\ 11\ 8) \neq (1\ 3\ 7) (11\ 8)$$ For ...

Top 50 recent answers are included