# Tag Info

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

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.

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.

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 ...

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 $n>2$.

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 $s = (2\ 3)(5\ 8)$ ...

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.

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 $G\le A_n$.

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 ...

2

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

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

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).

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.

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 $A_4$.

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 $X$ ...

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

First prove the following exercise: $$gHg^{-1}=g\mathrm{Stab}_G(a)g^{-1}=\mathrm{Stab}_G(g.a)$$ Since the action of $G$ is transitive, every $b\in A$ is equal to $g.a$ for some $g\in G$, so we have $$\cap_{g\in G}gHg^{-1}=\cap_{a\in A}\mathrm{Stab}(a)=:K$$ Now, $K$ is the set of elements $g\in G$ such that $g.a=a$ for all $a\in A$. But $G\leq S_A$, and the ...

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 $p \nmid (n-1)\dots(n-p+1)$.

1

For question 1. Take $f$ an element of $G$ acting trivially. Then for all $a$ in $A$, $f.a=a$. Now $f.a:=f(a)$ so that $f(a)=a$. This is true for all $a$ so $f$ is the identity function on $A$. So $f$ is indeed the neutral element of $S_A$. For the second question demonstrate then use the following result : When $G$ acts on a set $X$ then for all $g$ in $G$ ...

1

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

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

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

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: $$\sigma(1)\mapsto \sigma^{-1}\sigma(1)=1\mapsto2\mapsto\sigma(2).$$

1

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

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.

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 $S_n$?

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 $4!$?

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, ...

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 $n > 3$.

Only top voted, non community-wiki answers of a minimum length are eligible