G is a group and N a subgroup of G, and if index of N in G is 2, then N is Normal. Can someone please explain why?
123357
$\\$   A normal subgroup $N$ is a subgroup where the left cosets are the same as the right cosets. $N$ is normal $\iff $ $xnx^{-1} \in N,  \forall x\in G$.   5.) Why is it that if $[G:H]=2 \implies $ $H$ is normal subgroup?  6.) Can we say that a factor group is just a group that has left cosets of $N$ (being a normal subgroup) as its elements? So if $N$ is a normal subgroup, then the left cosets of $N$ forms a group under coset multiplication given by $aNbN = abN$.   7.) The group of left cosets of $N$ in $G$ is called the factor group, why do we denote this by $G/N$? These are the same things as the integers modulo $n$ groups? How can I relate those exactly?
91337
I am attempting to prove that, given a non-trivial normal subgroup $N$ of a finite group $G$, we have that $G$ is solvable iff both $N$, $G/N$ are solvable. I was able to show that if $N,G/N$ are solvable, then $G$ is; also, that if $G$ is solvable, then $G/N$ is. I am stuck showing that $N$ must be solvable if $G$ is.  It seems intuitive that any subgroup of a finite solvable group is necessarily solvable, as well. Is this true in general? For normal subgroups? How can I go about proving this result?  Edit: By solvable, I mean we have a finite sequence $1=G_0\unlhd ... \unlhd G_k=G$ such that $G_{j+1}/G_j$ is abelian for each $1\leq j&lt;k$.
109521
Let $p$ be a prime number and $G$ a finite group where $|G|=p^n$, $n \in \mathbb{Z_+}$. Show that any subgroup of index $p$ in it is normal in $G$. Conclude that any group of order $p^2$ have a normal subgroup of order $p$, but without using the Sylow theorems.
44913
This is , I think an easy problem just that I am not getting the catch of it. How to show whether or not the statement is true?  All subgroups of a group are normal$\implies$ the group is an abelian group?  I have been able to show the other way round.
1594
Let $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$. Is $M$ normal subgroup in $N$? Why or why not?  I know how to do this with $GL_2(\mathbb{R})$ and $SL_2(\mathbb{R})$ but with $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$ I don't even know all of their elements so I can't check the left and right cosets of $M$ in $N$.
69766
Let $G$ be a finite group with the property that all of its proper subgroups are abelian. Let $N$ be a normal subgroup of $G$. Prove that either $N$ is contained in the center of $G$ or else $G$ has a normal abelian subgroup of prime index.  I think $G$ is solvable. . I hope that idea maybe usefull.  Help me some hints.  Thanks a lot.  P/s: This is a question comes from a qualifying exam in Algebra ( Wisconsin August $1979$ )
97054
If G is a finite group and N is a normal subgroup in G , then can we say G $\cong$ G/N $\times$ N always? Is it true for like normal nilpotent or normal solvable or any such special classes.   I couldn't help but ask for it. What if G is infinite? Please tell me if you can, the concept I am skipping here. If you can avoid using exact sequences, It will be much clearer to me. I have seen an answer which still is not so clear to me. Thanks!
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.
57017
Let $G$ be a finite group and $H$ a subgroup of index $p$, where $p$ is a prime. If $\operatorname{gcd}(|H|, p-1)=1$, then $H$ must be normal. Does somebody have a quick proof of this?
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.
75341
Question is that :  let $N\unlhd G$ such that every subgroup of $N$ is Normal in $G$ and  $C_G(N)\subset N$.   Prove that $G/N$ is abelian.  what could be the possible first thought (though for me it took some time :)) is to use that $C_G(N)$ is Normal subgroup (As in general centralizer is a subgroup). one reason to see this is that $C_G(N)$ is not Normal in General and $C_G(N)$ is not subset of $N$ in general.  As $C_G(N)\subset N$, we have $G/N\leq G/C_G(N)$   I some how want to say that $G/C_G(N)$ is abelian and by that conclude that $G/N$ is abelian.  I would like someone to see If my way of approach is correct/simple??   I have not yet proved that $G/C_G(N)$ is abelian. I would be thankful if someone can give an idea.  Thank You.
51723
Suppose $H$ is the only subgroup of order $o(H)$ in the finite group $G$. Prove that $H$ is a normal subgroup of $G$.  I've been trying this problem for quite a while but to no avail. What I can't understand is, how do you relate the subgroup being normal/abnormal to its order?  This question is from I.N.Herstein's book Topics in Algebra Page 53, Problem no. 9. This is NOT a homework problem!! I'm studying this book on my own.
91271
Let $N$ be a subgroup of a group $G$. Suppose that, for each $a$ in $G$, there exists $a, b$ in $G$ such that $Na=bN$. Prove that $N$ is a normal subgroup.   Attack: I found $b^{-1}N$ = $Na^{-1}$ but I am stuck!  Any help will be appreciated
29098
Prove that if a group is containing no subgroup of index 2 then any subgroup of index 3 is normal.    Thank you.
67287
Let $G$ be a group and $H$ be a subgroup of $G$ with finite index. I want to show that there exists a normal subgroup $N$ of $G$ with finite index and $N \subset H$. The hint for this exercise is to find a homomorphism $G \to S_n$ for $n := [G:H]$ with kernel contained in $H$.   The standard solution suggests to choose $\varphi$ as the homomorphism induced by left-multiplication $\varphi: G \to S(G/H) \cong S_n$. I'm not 100% sure if I understand this correctly. What exactly does $\varphi$ do? We take $g \in G$ and send it to a bijection $\varphi_g: G/H \to G/H, xH \mapsto gxH$? If so, how can I see that its kernel is contained in $H$? Also, the standard solution claims its image is isomorphic to $G/N$ and thus $N$ has a finite index in $G$, how can I see that the image is isomorphic to $G/N$?  Thanks in advance for any help.
58605
Question is:     Let $G$ be a group of  order 2n. Suppose half of the element of G are of order 2 and the other half form  a subgroup $H$ of order n . Prove that $H$ is of odd order  and is an abelian subgroup of $G$   What could i see is..  if we prove that order of every element of $H$ is odd , then order of $H$ is odd . Also i am unable to use the fact that half of the element of $G$ are of order 2.  Please help me to clear this.  Thank You.
103202
I've studied quite a bit of group theory recently, but I'm still not able to grok why normal subgroups are so important, to the extent that theorems like $(G/H)/(K/H)\approx G/K$ don't hold unless $K$ is normal, or that short exact sequences $1\to N \stackrel{f}{\to}G\stackrel{g}{\to}H\to1$ only holds when $N$ is normal.   Is there a fundamental feature of the structure of normal subgroups that makes things that only apply to normal subgroups crop up so profusely in group theory?  I'm looking here for something a bit more than "$gN=Ng$, so it acts nicely".