Let $G$ be à profinite group and $\widehat{G}$ his profinite completion. I want to show (I hope it's true) that $\widehat{G}$ is strongly complete that is all normal subgroups of finite index are open sets.  For that I take $\widehat{N}$ a finite index normal subgroup of $\widehat{G}$ and try to find $N$ normal subgroup of finite index of $G$ such that $\widehat{N}=\ker\left(\varphi_N:\widehat{G}\to G/N\right)$. I tried $N=\theta^{-1}\left(\widehat{N}\right)$ where $\theta:G\to\widehat{G}$ but could not conclude.  Not rely with  I think: I don't want a different definition of strongly complete but I want to check that somewath ($\widehat{G}$ here) is strongly complete.
122144
Let $G$ be a profinite group (or equivalently a  compact and totally disconnected topological group ) with the property that all of its normal subgroups of finite index are open sets.   Does this imply that all of its subgroups of finite index are open sets ?  (if all subgroups of finite index from $G$ are open sets, than $G$ is called strongly complete ; this motivates the title of this post)
12270
I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.   My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.   I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).   First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?  Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l $. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).   Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine.
115013
Let $G$ be a group of order $150$.  I must show that it has a normal subgroup of order $25$.  The hint says to show that is has a normal subgroup of order $5$ or $25$.  Now from Sylow, I know that the number $n_5$ of Sylow-$5$ subgroups (which each have $25$ elements) must be either $1$ or $6$, since $n_5$ must also divide $6$.  Now clearly if $n_5=1$ we are done, so I can assume that $n_5=6$.  My problem is I haven't figured out how to use this information (I am going for a contradiction, just based on what the problem asks me to prove).  I know by Cauchy's theorem I can get a subgroup of order $5$, but I don't see immediately if it must be normal.   Any direction I should try to be moving?
73376
I'm stuck on the following practice problem.   Any hints would be appreciated.  Suppose $N$ is a normal subgroup of  $G$ such that every subgroup of $N$ is normal in $G$ and $C_{G}(N) \subset N$. Prove that $G/N$ is abelian.  I'm not sure how to use the fact that $C_{G}(N) \subset N$.  Thanks
106839
My question is simple:  Is the category of profinite groups an accessible category?  Thank you  Edit: I will add the (hopefully simpler) question:  Is the category of profinite groups complete and does it admit a cogenerator?
18714
Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.  Can somebody help me out?
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.
43088
I have a finite $p$-group $G$ and a normal subgroup $N$ which is not the trivial subgroup. I am asked to show that $|N \cap Z(G)| &gt; 1$.  There has been a similar question on MSE here:  Unfortunately, I cannot comment on other user's questions yet, so I made my own question.  I know that the trick is to use conjugation but I can't follow the logic. Can someone elaborate? Especially on the last part of the top answer.
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.
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.
114783
I'm trying to prove that there exists an open neighborhood of the identity in the complex general linear group that contains no proper subgroup.  This is needed to show that a continuous representation of a profinite group has open kernel.
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...)
38487
Let $K=\{(1),(1,2),(3,4),(2,4),(1,4),(2,3)\}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.  I know that i have to examine the condition of normality in $S_4$ i.e. for all $g\in S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.  Edit: $K = \{(1),(12)(34),(13)(24),(14)(23)\}.$
70921
Prove that if H has finite index n then there is a normal subgroup N of G with $N \subset H $ and [G:N]$ \le $n!.  I tried to solve the problem but could not done exactly. Since [G:H]=n , Let A={$a_i H$:i=1,2,..,n}. Consider the left multiplication group action on A .Then kernal of the action is intersection N= $\cap a_iHa_i^{-1}$ i=1,2..,n.  Then N is the largest normal subgroup of G contained in  H. Now it is clear that $[G:a_iHa_i^{-1}]$ =n for all i.{by orbit stabliser theorem}.Now how to find [G:N]? yes I got it. By Caley's theorem G/N is isomorphic to a subgroup of $ S_{A}$ .Therefone [G:N] divides n!.
24728
I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example?
122260
I'm trying to prove if $|G|=p^nq$ with $p\gt q$, primes, then $G$ contains a unique normal subgroup of index $q$.  I know by the first Sylow theorem that G has a Sylow p-subgroup $P$ with $[G:P]=q$. My problem is prove that it's unique.  I need help  Thanks in advance
102548
Let $G$ be a finite abelian group and let $\phi: G \rightarrow G$ be a group homomorphism. I am trying to show that there is a positive integer $n$ such that $G \cong \ker(\phi^{n}) \times \phi^{n}(G)$.  I know that since $G$ is abelian we have that $\ker(\phi^{k})$ and $\phi^{k}(G)$ are normal subgroups of $G$ for any $k$  I also suspect we have the following towers which stabilize at $m$ and $m'$  $$\phi(G) \unrhd \phi^{2}(G) \unrhd \cdots \unrhd \phi^{m}(G)=\phi^{m+1}(G)=\cdots$$  $$\ker(\phi) \unlhd \ker^{2}(\phi) \unlhd \cdots \unlhd \ker^{m'}(\phi)=\ker^{m'+1}(\phi)=\cdots$$  I know that if given a group of the form $HK$ where $H$ and $K$ are normal subgroups of $HK$ and $H\cap K=1$ then $H \times K \cong HK$. I want to apply this to this situation but I am unable to show how write $G$ as a product $HK$  Resolution One of the comments directed me to the following article:  The last part of which answers the question. Below is what is says:  Choose $n=\max(m,m')$, then we have for $x \in \ker^{n}(\phi) \cap \phi^{n}(x)$, this means that $x=\phi^{n}(y)$ for some $y \in G$. This gives:  $0=\phi^{n}(x)=\phi^{2n}(y)$ which means that $y \in \ker^{2n}\phi=\ker^{n}{\phi}$. and then we have that $0=x=\phi^{n}(y)$.  The answer below kindly points out that every element $x$ is contained in one of the cosets of $G/\ker^{n}(\phi)$, this means that $x=k+g$ for some $k \in \ker^{n}(\phi)$ and $g \in G$. We also have that $g=\phi^{n}(h)$ for some $h \in G$. Writing $x=k+\phi^{n}(h)$ essentially shows that $G=\ker(\phi^{n}) + \phi^{n}(G)$. Using the fact above, the claim follows.