The following questions were suggested by my friend, while we were studying fundamental group theory. We had no exact ideas of the way to approach the problems. Questions  (1) Let $G$ and $H$ be groups such that $|H|=|G|$. If we can make a bijection $\phi:G\to H$ such that for $\forall a\in G$, $ord(a)=ord(\phi(a))$, then does $G\cong H$?   (2) Let $G$ be a group. If all of the elements in $G$ except $e$ have the same order, then is $G$ an abelian group?  Both suggestions were not easy to find the counter-examples. So, we wanted to prove these statements instead, but again too difficult. We also failed to find these types of questions on the internet. Is there theories or propositions related to the problems? Thanks.
11401
Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.
18939
Let $H \leq G$. If $G = H H^x$ then $G=H$.  According to the source, it should solve with elementary knowledge of group theory, but my tries have no result! Any help would be appreciated.   Thanks in advance.
63557
Aluffi II.3.4 asks the reader to find groups $G$ and $H$ such that $G\times H\cong G$ but $H$ is not trivial. I believe I have found a solution, but would like someone to check. Also, is there a more elementary example? I tried mucking around with $S^1\times (\Bbb Z/2\Bbb Z)$, which feels like it should be isomorphic to $S^1$ (the unit circle under complex multiplication), but I kept getting confused.  My solution  Let $H$ be a non-trivial group and let $G=H^\omega$. Then $\phi\colon H\times G\to G$ defined by $$\phi(h,g)_n=\begin{cases} h &amp;n=0\\ g_{n-1} &amp;n\ne 0 \end{cases}$$ looks like a group isomorphism to me. It's obviously a bijective function, and \begin{align*} \phi((h_1,g_1)(h_2,g_2))_n &amp;=\phi(h_1h_2,g_1g_2)_n \\ &amp;=\begin{cases} h_1h_2&amp;n=0\\ (g_1g_2)_{n-1}&amp;n\ne 0 \end{cases}\\ &amp;=\phi(h_1,g_1)_n\phi(h_2,g_2)_n \\ &amp;=(\phi(h_1,g_1)\phi(h_2,g_2))_n. \end{align*}
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?
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?
5675
I have the following question:      Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$   Then it holds: $g^{n!}\in H$ for all $g\in G$.   Why is this true?  I think, that's not very difficult, but I have no idea at the moment.   Thanks!
118745
Let $f:G\to K$ be a surjective group homomorphism and $H$ be a subgroup of $G$ satisfying $N_G(H)=H$. Then $N_K(f(H))=f(H)$. I'm trying to prove or disprove it. perhaps, this proposition is true. But, how can we prove it? Edit: I thought it is true because WIKI said it is true without proof
69340
Question:  Give an example of a group $G$ having a subset $H$ such that $HH=H$, where $HH = \left\{h_1h_2\mid h_1,h_2\in H \right\}$, but $H$ is not a subgroup of $G$.   Attempted Solution:  I haven't gotten far with this problem. In a previous exercise, I was able to prove that $HH=H$ if $H$ is a subgroup of $G$. I am having a hard time thinking of a scenario where $h_1h_2\in H$ and $H$ is not a group. I've played around with making $h_1h_2$ not closed under its operation, but I haven't gotten anywhere yet.
89495
Let $G$ be a group and $H$ a subgroup with $[G:H]=n$. Is it true that $x^n\in H$ for all $x\in G$?   Remarks. The answer is positive whenever $H$ is normal, e.g., for $n=2$. In general, by using the normal core of $H$, one can find an $m\ge 1$ such that $x^m\in H$ for all $x\in G$.
42433
I'm trying to figure out this problem:  Problem: Let $G$ be a group of order $p^k q$ with $p$ and $q$ distinct primes and $k \in \mathbb{N} \setminus \left\{0\right\}$.   a) Prove: $G$ has a subgroup of order $p^k$, or $p$ divides $|Z(G)|$. (Here $Z(G)$ means the center of $G$).   b) Prove: if $p$ divides $| Z(G)|$, there there exists an $x \in Z(G)$ of order $p$.   Attempt: For a) I tried proceeding by induction.  Base step: Assume $k=1$. Then $|G| = pq$ with $p \neq q$. Case 1: if $G$ is commutative, then $Z(G) = G$ and then $p$ divides $|Z(G)|$.   Case 2: Assume $G$ is not commutative. Then the center of $G$ is trivial, i.e. $Z(G) = \left\{e_G\right\}$ (this is a theorem of group theory). So $p$ does not divide $|Z(G)|$. Then we must show that $G$ has a subgroup of order $p$. Let $H$ be a non-trivial subgroup of $G$, that is $H \neq G$ and $H \neq \left\{e\right\}$. Then by Lagrange, $|H|$ divides $pq$. So we must have either that $|H| = p$ or $|H| = q$ (is this correct? Since $H$ is non-trivial). If $|H| = p$ then we are done. If $| H | = q$, then $|G / H| = p$ and so $G/H$ is cyclic. I want to derive a contradiction from this, and so show that $H$ cannot have order $q$, but I don't know how.  Induction step: Assume the claim holds for $|G| = p^{k-1} q$. We must then show that it also holds for $|G| = p^k q$.  I want to use the class equation. We have that $$ p^k q = |Z(G)| + \sum_{i=m+1}^n |Cl(a_i)| $$ where $|Cl(a_1)| = \ldots = |Cl(a_m)| = 1$ and $|Cl(a_i)| &gt; 1$ for $i &gt; m $. Since $p$ is a divisor of the LHS, it must also divide the linear combination. But how can I conclude from this that $p$ divides $| Z(G)|$ ? I don't know how to use the induction hypothesis.   b) Assume $p$ divides $|Z(G)|$. I first assume that $G$ is abelian. Then by structure theorem for finite abelian groups, we know that $$ G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \ldots \oplus \mathbb{Z}_{p_r^{k_r}}. $$ This means that $$ p^k q = p_1^{k_1} \ldots p_r^{k_r}. $$ So $p | (p_1^{k_1} \ldots p_r^{k_r})$ and so since $p$ is prime, $p$ must divide at least one factor. Assume that $p$ divides $p_i^{k_i}$. Since $p_i^{k_i} = p_i \ldots p_i$ with $k_i$-amount of factors, we must again have that $p$ divides $p_i$. So $p = p_i$.   Now, since every $\mathbb{Z}_{p_i^{k_i}} = \mathbb{Z}_p^{k_i}$ is abelian, we have that the center equals the whole group. Let $x_i$ be an element in this center. Then $order(x_i) | p^{k_i}$ and so $order(x_i) | p$. Hence the order of $x_i$ equals $p$ in $\mathbb{Z}_p^{k_i}$. Since $x_i$ is also in $Z(G)$, this concludes the proof?   And what about the case that $G$ is not abelian?  Help with this problem is appreciated!
103432
Just starting to properly learn the chapter on homomorphisms, and it will be great if I can understand the following question I have:  How could I actually show that when there are two groups $G$ and $H$ (finite), and $|G|$ and $|H|$ are coprime, then then homomorphism $\Phi:G \rightarrow H$ must be trivial.   My idea: Since $Im(\phi)$ and $Ker(\phi)$ are respective subgroups of the groups $G$ and $H$, then by Lagrange's Theorem $|Ker(\phi)|||G|$ and $|Im(\phi)|||H|$. I think if we just show that the |$Im(\phi)$| is $1$, then it implies that the $|Ker(\phi)|$ is equal to the order of $G$. Hence $\phi$ must be trivial. But if this is correct, I cant seem to show $|Im(\phi)|=1$. However, is this the correct way?   Can someone please help me show the homomorphism $\Phi$ is trivial if my idea is completely off?
97103
$(G,*)$ is an abelian group and $a,b$ are elements of G. Let $n=ord(a)$ and $m=ord(b)$ in $G$. Show that the order of $a*b$ is equal to $nm$ if $gcd(n,m)=1$.  I have already proved that $ord(a*b)|nm$. But the other part is a problem.  Thanks
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.
32086
Let $G$ be any group, and $H \leq G$ a subgroup.  Suppose that for each $x \in G$, there exists a $y \in G$ such that $xH \subseteq Hy$. In other words, every left coset of $H$ is contained inside some right coset of $H$.  Question: what can we say about $H$? In particular, does this imply that $H$ is normal?  I know that if $G$ is finite, then $H$ must be normal, because then $xH \subseteq Hy$ implies $xH = Hy$, since $|xH| = |Hy|$.
85040
Given the following definition of normal subgroup     A subgroup $H$ of a group $G$ is said to be normal if, for every $g\in G$: $$gH=Hg$$   I've tried to show that $H\mathrel{\unlhd} G$ if and only if we have $gHg^{-1}=H$ for every $g\in G$:  (if) Let $H\mathrel{\unlhd} G$, $g\in G$; if $l\in gHg^{-1}$, there exists $h\in H$ such that $l=ghg^{-1}$ and since $gh$ = $h'g$ for some $h'\in H$, then $l=ghg^{-1}=h'(gg^{-1}) = h'\in H$. Let $h\in H$; we can write $h$ as $hgg^{-1}$, and since $hg = gh'$ for some $h'\in H$, we obtain $h = hgg^{-1} = gh'g^{-1}\in ghg^{-1}$. So $gHg^{-1}=H$.  (only if) Let $gHg^{-1}=H$, $g\in G$; if we take an $l\in Hg$, $l = hg$ for some $h\in H$, we note immediately that $l=hg=(gh'g^{-1})g=gh'\in gH$. What should I do to show that $gH\subset Hg$?  I'm also wondering if this mess is correct, and if it is possible a more elegant proof of this fact.
71935
When people want to classify a group with certain (small) order, they seem to find a normal subgroup $H$ and a subgroup $K$, and then consider how many distinct $H \rtimes_\varphi K$ are there.  My question is: For any group $G$, does it always exist normal subgroup $H$, subgroup $K$ and $\phi : K \rightarrow Aut(H)$ s.t. $G\cong H \rtimes_\varphi K $?  Sorry if this question is too basic...  Thanks the comment and the answers. Then if the $G$ is not simple, can we have can $G\cong H\rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal?  EDIT:  If we have a group $G$, that is not simple (then $G$ has at least one normal subgroup), are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible).  But the quaternion group and $Z_4$ given in answers have shown that we cannot always have it.
30957
In a question from an exam in an undergraduate group theory course, we were asked to prove or disprove that the set of all Torsion elements of a group is necessarily a subgroup.  I knew that the set of Torsion elements is closed under the inverse operation, but was later told that it is not closed under multiplication, therefore disproving the claim. However, I couldn't find any examples of a group $G$ and two elements $a,b$ such that both $a$ and $b$ have a finite order, but $ab$ doesn't. I know that for this to happen $G$ must be an infinite and non-Abelian group, but still couldn't find a valid example.   What are some examples of groups/elements fulfilling the aforementioned property?