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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 ...


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. ...


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:


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 ...


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

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 $3$ ...


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.


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 ...


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

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 ...



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