Let $G$ be a finite group and $H$ be a subgroup of $G$. Show that if $aH$ intersect $bH$ is not empty, then $aH$ and $bH$ contain an equal number of elements in $G$. Conclude the number of elements in $H$ divides the number of elements in $G$.
123267
I'm trying to get some intuition on why order of a subgroup divides order of a group. I know Lagrange's Theorem and its proof (every element of a group is an element of some coest, cosets are disjoint and each one has the same number of elements as subgroup), but the proof is more like magic than any intuition.  So I've been trying to prove this fact (divisibility of order) without Lagrange theorem. Let's say we take a group $H$ and try to extend it, by taking $g \notin H$ and considering group $G$ generated by set $H \cup {g}$. Such a group would have to contain $g^{-1}$ and every word over alphabet $\{g, g^{-1}, H\}$. Is it possible to prove intuitively that such construct is of size $|H|\cdot n$ for some $n \in \mathbb{N}$? This would essentially prove that any group $G$ containing $H$ has order divisible by $|H$|, because we would be able to take $g \in G$ such that $g \notin H$, extend $H$ by it just like above and continue this process until $H$ is extended into $G$.
41242
I need help to prove this result:  "Let $G$ be a group such that the intersection of all its subgroups other than $\{1\}$ is a subgroup different from $\{1\}$. Then all its elements have a finite order".  I know I must think of an element $g$ of infinite order. That will imply that every subgroup has infinite order (because it will contain an element like $g^k$). After this, I don't know which step I can take. Can someone help?
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.
107944
I have been thinking about this problem:  "Suppose that a group $G$ has a subgroup of order $n$.  Prove that the intersection of all subgroups of $G$ of order $n$ is a normal subgroup of $G$."  Unfortunately, I don't have much of a start.  I know that the intersection will be a subgroup of each of the subgroups of order $n$, and thus will have order dividing $n$.  I also know that conjugation preserves the order of elements, which suggests that $xNx^{-1}$ could be a subset of $N$ ($N$ being the intersection of all subgroups of order $n$), but I don't see why conjugation necessarily takes elements of $N$ to other elements of $N$.  I'd really appreciate a hint on how to go about showing this.  Thanks.
96242
Say I have a finite group and I tell you how many elements and how many subgroups it has. Do you now know what group it is? Or could two groups which are not isomorphic to each other have the same amount of elements and subgroups? In other words, does the amount of elements and subgroups of a group define it completely? It is interesting because if I give you just the number of elements there are several possibilities and if I give you the number of subgroups also but perhaps both combined are enough to define the group.
19200
Question. Let $G$ be a finite group and $p$ be the smallest prime dividing $|G|$.  Let $x$ be n element of order $p$ in $G$. Assume that there exists an element $h\in G$ such that $hxh^{-1}=x^{10}$.  Show that $p=3$.    What I have done so far is use the fact that $o(hxh^{-1})=o(x)=p$ and $o(x^{10})=p/\gcd(p,10)$.  By hypothesis we have $p=p/\gcd(p,10)$, giving $\gcd(p,10)=1$.  Here I am stuck.  I have not been able to make use of the fact that $p$ is the smallest prime divisor of $|G|$.  Can somebody help?
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.
40857
Let G be a finite group whose order is not divisible by $3$. Show that for every $g\in G$ there exists an $h\in G$ such that $g = h^3$.
5791
I'm looking for verification:  My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k&gt;1$, where $k! \leq |G|$, then $G$ is not simple.  Proof: Consider the set of left cosets of $G$ by $H$:  $$G/H = \{H,g_1H,\dots,g_kH\}$$  Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:  $$\phi(g)[g_iH] = g g_iH$$  ($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)  Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.  Because $\phi$ is a homomorphism,  $$ G/\ker\phi \cong \text{im}\phi$$  Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| &gt; 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.  Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.
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
27055
*****************updated question**************  The question starts off like this:   Let $H$ be a subgroup of a finite group $G$. Prove that $|Ha| = |Hb|$ by proving that there exists a bijection $f : Ha \rightarrow Hb$. Find this bijection.   I know that I have to show injective and surjective but want to know if anyone thinks this is correct in this fashion  $$Let\ a\in G, \ y, x \in Ha,then \ x=ra, y=sa\ for\ some\ r,s \in H. $$ $$we\ define\ f:Ha \rightarrow Hb\ by \ f(x)=hb$$ $$suppose \ f(x)=f(y)$$ $$rb=sb\ for\ some\ r,s \in H$$   $$r=s$$   $$xa^{-1}=ya^{-1}\ (from\ x=ra\ and\ y=sa)$$ $$x=y$$ This shows injective. for the onto part this is really wanted to check $$let\ x\in Ha, y\in Hb\ and\ a,b \in G\  such\ that$$ $$x=sa,\ y=sb\ for \ some\ s \in H$$ $$then\ f(x)=sb$$ $$=yb^{-1}b\ (from \ y=sb)$$ $$=y$$ Thus, we have shown that $\forall y \in Hb, \exists\ x \in Ha \ such \ that f(x)=y  $  here we have shown the function is also surjective and hence a bijection exist between $Ha$ and $Hb$ and therefore we know then  $|Ha|=|Hb|$
6113
Suppose we have a set of $n$ elements. We want to know the maximum number of subsets of $3$ elements we can find of this set such that no two subsets have an intersection with $2$ or more elements.  The question arises from trying to count the number of subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ in a group with $n$ elements of order $2$ such that the product of any two such elements has also order $2$.
120710
The problem I am attempting to prove is the following: In any finite group $G$, the number of elements not equal to their own inverse is an even number.  Caveat: I have had very limited experience with proofs. Any comments would be appreciated.  Let $a_{1}, a_{2}, ..., a_{n}$ be elements of $G$. Since $G$ is a group, every element in $G$ must have an inverse in $G$. Let the inverse of each element be represented by $a_{1}^{-1}, a_{2}^{-1}, ..., a_{n}^{-1}$. An element $a_{k}$ in $G$ that is not equal to its own inverse can be written as $a_{k} \neq a_{k}^{-1}$. For every element $a_{k}$, there is another element $a_{k}^{-1}$ that is not equal to its own inverse denoted by $a_{k}^{-1} \neq a_{k}$. So for every element $a_{k}$ in $G$ whose inverse is not equal to their own inverse, there will be 2 elements in $G$ namely $a_{k}$ and $a_{k}^{-1}$. Let the integer $q$ denote the number of elements ranging from $a_{1}, a_{2}, ..., a_{n}$ who are not equal to their own inverse. In addition for every value $q$, there must be another element in $a_{1}^{-1}, a_{2}^{-1}, ..., a_{n}^{-1}$ not equal to their own inverse. So the total number of elements not equal to their own inverse can be denoted by 2$q$, which is an even number.
105562
Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$  Question: Deduce that if $G$ is a finite $p$-group, for some prime $p$, the number of subgroups of $G$ that are not normal is divisible by $p$.  Comments:    The subgroups of $G$ that are normal have the property that $g\cdot H=gHg^{-1}=H$ The question is equivalent to showing $p$ divides $\left|G\right|-\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$ $p$-group: $\forall g\in G,|g|=p^k$ for $k\in\mathbb{N}^+$ I don't know where to start with this problem, been thinking about it for a while to no avail, I feel that there are too many definitions for me to consider when finding a solution. My problems I have considered and not been able to answer are: what is the order of $|G|$? I think it must be of order that is divisible by $p$, hence the question becomes show $p$ divides the order of $\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$. How do you compute the order of this set? It seems like quite a standard problem, so I would really appreciate it if I could be directed to more information about whatever problem it is.
88763
Let G be a group with $o(G)=pq$, $p&gt;q$ are primes.  Prove that $G$ has a subgroup of order $p$ and a subgroup of order $q$.  Remark: I cannot find the topic with this problem.  I have tried different methods but no one of them did not bring result. Please show how to do that. Do not use Sylow's Theorem.
95758
Let $G$ be a group, and let $H$ be a subgroup of $G$. Let $x,y\in G$ be such that $xH\subseteq Hy$. Can we conclude that $xH=Hy$? this certainly holds if $H$ is finite because both sets will have the same finite cardinality. But what about the case $|H|=\infty$? I am especially interested in the case $x=y$.  EDIT: As Thomas Andrews pointed out in the comments, it is suffices to consider the case $x=y$, because the hypothesis $xH\subseteq Hy$ implies $x\in Hy$, hence $Hy=Hx$. And so, the question becomes:     Can a subgroup (of a group, of course) strictly contain some of its conjugates?
52607
Finite group $G$ contains elements $g_1,g_2,...,g_n$. Show that $g_1g,g_2g,...,g_ng$ consists of every element from $G$, where $g\in G$.  I know that the operation of a finite group gives back each element of the group once, but how can I prove this? I know I have to show that some $g_ig$ is in the group and some $g_ig$ does not equal some $g_jg$