If $H$ is a subgroup of $S_n$ and if H is not contained in $A_n$ prove that precisely one half of the elements of H are even permutations. I know that multiplying two odd permutations gives a even, and that one even and one odd return an odd permutation but i have no idea where to go from there.
122837
Let $H&lt;S_{n}$ and suppose $H$ is not contained in $A_{n}$. Write $H$ as $$H=E\cup O$$ where $E$ and $O$ represent the sets of even and odd elements, respectively. Let $E=\{{\alpha_1,...,\alpha_n}\}$ and $O=\{{\beta_1,...,\beta_m}\}$. Since $O$ is non-empty, it contains at least one element, say $\beta_1$. The same goes for $E$ because $e\in E$. Our goal is to prove $m=n$. The products $\beta_1\alpha_i$, $i=1,...n$, are all distinct elements in $O$, and it follows that $O$ has at least $n$ elements, i.e. $m\geq n$. In the case of strict inequality, $m&gt;n$, add at least one extra odd element ($\beta_{n+1}$) to the list $$\beta_1\alpha_1,...,\beta_1\alpha_n,\beta_{n+1}$$  This is where I find myself stuck. Any help appreciated. Can this route work?
20677
In , people discussed that there cannot be cyclics subgroups of order 9 in $S_6$.   How does one reason that the subgroups isomorphic to $C_3 \times C_3$ (the other candidates to be subgroups of order 9) are precisely the ones given by two disjoint permutations $\langle (123),(456) \rangle$
95884
Given that $S_3$ is a symmetric group of size three, how would I find all elements of it, and all subgroups?
104817
Let $G$ be a finite group and let $H$ be a subgroup of $G$ with the property that no other subgroup of $G$ has order $|H|$. Prove that $H$ is normal.  My attempt:   To show that $H$ is normal, I need to show that for any $g,h \in G$, $ghg^{-1}\in H$.  Let the order of $G=n$ and let the order of $H=k$. If $H$ is a subgroup of $G$ then $k$ divides $n$ and the identity element $e$ is both $G$ and $H$. Also for any $a,b \in H$, $ab^{-1}\in H$.  Since $k$ divides $n$, $n=qk$ for some integer $q$.  $a^{n}=a^{qk}=e$...  (I don't think this proof is getting me anywhere. I think I need a hint about what does "$H$ being a subgroup of $G$ such that no other subgroup of $G$ has order $|H|$" imply...)
63191
Prove the following in $S_n$:  Let $\alpha = (a_1,a_2,\ldots,a_s)$ be a cycle and let $\pi$ be a permutation in $S_n$. Then $\pi\alpha\pi^{-1}$ is the cycle $(\pi(a_1),\ldots,\pi(a_s)).$     I'm not sure where to even start.
73465
I have been going through some material on elementary group theory () and in one place it states:  A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.  Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$. How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.
54470
Let $G$ be a simple group and there exists a subgroup $H$ of index $n \in \mathbb{N}_{\ge 3}$. Show that: $|G|$ divides $n!/2.$ I saw a similar question .   There was written: "The proof is really simple. Let $X$ be the set of left cosets of $H$. Consider $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$. Then $\phi$ is a homomorphism. Consider now $K=\ker \phi$. Then $K$ is a normal subgroup of $G$ contained in $H$. Finally, $G/K$ is isomorphic to a subgroup of $\text{Sym}(X)$, which has order $n!$, where $n=[G:H]$. Thus, $[G:K]$ is finite and divides $n!$."  In fact, I don't understand the last sentence. "Thus, $[G:K]$ is finite and divides $n!$." Also did I understand it right that $ker(\phi)=H$?  Can you also give me an advice how to proof it for $n \ge 3$. (maybe induction)  Thank you for taking your time.
60423
If $H$ be a subgroup of a group $G$  such that $Ha \not=Hb$ implies that $aH\not=bH$.Then how can I show that $gHg^{-1}\subset H$  $\forall$ $g\in G$?  I do not think I have made any progress.However, this is what I have done: For any $h\in H$,$(ghg^{-1})(gh^{-1}g^{-1})=e$ where $e$ is the identity element of the group $G$.  And for $h,k\in H$, $(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$ which is in $gHg^{-1}$ as $hk\in H$. Thus $gHg^{-1}$ is a group.  I do not think I am getting anywhere.
69056
Let $H,K⊲G$ be two normal subgroups such that $H∩K = \{1_G\}$. Prove all elements in $H$ commute with all elements in $K$.  I have no idea how to do this.
122881
How many subgroups of order $n$ does $D_n$ have?  My work:  Since subgroups of $D_n$ are either cyclic or dihedral, the subgroups of order $n$ of $D_n$ are $\left&lt; r \right&gt;$ (cyclic) and $D_{n/2}$ (dihedral). However, this works only for $n$ is even and $n\geq 2$. I don't know how to do when $n$ is odd.
39758
Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.  So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle?   Reference: Fraleigh p. 103  Question 10.35 in A First Course in Abstract Algebra
29567
Suppose $G$ is finite and $G=H\cup K\cup L$ for proper subgroups $H,K,L$. Show that $|G:H|=|G:K|=|G:L|=2$.  What I did: so if some of $H,K,L$ is contained in another, then we have $G$ being a union of two proper subgroups, which is impossible due to another result. So none of $H,K,L$ is contained is each other. If some element $h$ belongs to only $H$ and $k$ belongs to only $K$, then its product $hk$ cannot be in $H$ or $K$, so must belong to only $L$.
13548
Let $G$ be a finite group having order $n$, prove that $n$ is odd if and only if for each $a\in G$, there is $b\in G$ such that $a=b^2$. I first assume $n$ is odd, then let $a$ be an element in $G$, since $a^n=e$, we have $$a=a^{n+1}=(a^{(n+1)/2})^2$$ which tells $a$ is the square of $a^{(n+1)/2}$.  To prove the converse, I use contraposition, which I assume $n$ is even, then I want to find $a\in G$ such that it is not a square. My intuition is, "assume" we can always find an element $a\in G$ such that the smallest positive integer $k$ satisfy $$a^k=e$$ is $k=n$. Then it is not true that $a$ can be expressed as a square, indeed if $a=b^2$ for some $b\in G$, then the smallest $k$ satisfy $b^k=e$ will be $k=2n$, but this contradicts the smallest $k$ cannot exceed $n$.  My problem is, is the assumption I made always true? If it is true then it should be in the very beginning of the textbook, but I have forgotton it. Any help?  Source: Apostol Analytic Number Theory.
52992
True or False. Every subgroup of a simple group is itself simple.  Our solution:The statement is false.  As a counter example, let $G$ be equal to the group of even permutation of $10$ letters and $H$ is the subgroup of $G$ which is $$H=\{e,(12)(34),(56)(78),(12)(34)(56)(78)\}.$$ $H$ is not simple since $K=\{e,(12)(34)\}$ is also normal in $H$.
108820
Does every non-trivial subgroup of $S_9$ containing an odd permutation necessarily contain a transposition?    Here $S_9$ denotes the group of all permutations (i.e. bijections with itself) of the set $\{1,2, \ldots, 9 \}$ under the binary operation of composition of functions.   By a transposition is meant a permutation that leaves all elements except two fixed, mapping either of these two onto the other.   A permutation can only be expressed either as a product of an even number of transpositions or as a product of an odd number of transpositions (but not both) and as such is defined to be even or odd, respectively.   Reference: Fraleigh p. 95  Question 23d. in A First Course in Abstract Algebra
28705
The subgroup in $S_4$ that I know has order 12 is the subgroup of all even permutations, otherwise known as the alternating group $A_4$. However, I know this from a fact and not because I am able to show a subgroup of order 12 exists in $S_4$ in the first place. If I had not been told there existed a subgroup of order 12 in $S_4$, I would not have known there was one. So I guess this is actually a two-parter for me: 1) How do I go about showing that a subgroup of order 12 does indeed exist in $S_4$? I know by Lagrange that if there exists a subgroup, then the order of that subgroup must divide the order of the group. However, this doesn't say anything about the existence of the subgroup. And Sylow only verifies subgroups of a prime to some power order, which 12 is not. I also know that there might be a cyclic subgroup of order 12, but without manually multiplying every possible permutation in $S_4$, I don't know any other way to check the existence of this or if this subgroup is isomorphic to $A_4$ because I don't know how to write these permutations as functions in order to check if there is a homomorphism (I know that a permutation is bijective by definition). And 2) How do I show that this group of order 12 is the only group of order 12 in $S_4$?
112790
Im having trouble with the second part of this question,  Let $H$ be a normal subgroup of $G$ with $|G:H| = n$,  i) Prove $g^n \in H$ $\forall g \in G$ (which i have done)  ii) Give an example to show that this all false when $H$ is not normal in $G$.(which I am having trouble with showing)  Any suggestions?