I'm trying to show that any subgroup $H$ of a group $G$ having finite index must contain a normal subgroup of $G$ of finite index.  I tried to define a homomorphism $\psi:G/H \to G/H$ given by $\psi(xH) =gxH$ and prove that $\ker(\psi)\subset H$. Is there another way to solve this problem?
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.
54778
Reference:   Let $G$ be a group. Let $H,K$ be subgroups of $G$. How do I prove that $[G:H\cap K]\leq [G:H][G:K]$?  Let's not assume any index is finite. Then, still the result holds? If so how do I prove it?
14362
I am currently working my way through The Theory of Finite Groups by Kurzweil and Stellmacher and came across the following question in the first chapter:  Let $G$ be simple, $|G| \ne 2$, and $f : G \to H$ be a homomorphism. If $A \unlhd H$ is a normal subgroup of index $2$, then $\text{range}(f) \le A$.  Please could someone help me to prove this statement? It doesn't seem like it should be too difficult but having spent some time on it I'm still getting nowhere.  Thanks!
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.
49386
Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?  Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.  If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
105759
Prove or disprove: if $f:G \rightarrow H$ is a surjective group homomorphism, then   1) $f(A)$ normal implies $A$ normal.  2) $S$ is a p-Sylow subgroup of $G$ implies $f(S)$ is a Sylow-p subgroup of $H$.   I believe 1 is false since I believe we need injectivity, but I cannot think of a counterexample...  I believe 2 is true but do not know how to show it...
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
39278
Let $G$ be any group such that  $$G\cong G/H$$ where $H$ is a normal subgroup of $G$.  If $G$ is finite, then $H$ is the trivial subgroup $\{e\}$. Does the result still hold when $G$ is infinite ? In what kind of group could I search for a counterexample ?
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?
116253
Let $\varphi:G\rightarrow H$ be a group homomorphism from group $G$ to group $H$. Show that, if $H$ is abelian, all subgroups of $G$ that contain $\mathrm{ker} (\varphi)$ are normal in $G$.
4496
Let $G$ be a group, $H\subseteq G$ a subgroup and $a\in G$ an element of the group. Is it possible that $aHa^{-1} \subset H$, but $aHa^{-1} \neq H$?   If $H$ has finite index or finite order, this is not possible.
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 $1979$ )
109600
I was having a discussion with a friend of mine about some normal group properties and then came up with the question "if G is not commutative, then is there always a subgroup that is not a normal subgroup?" It's probably more easy to solve this in the following form:$$\forall H \leq G : H  \lhd G \Rightarrow \forall a,b \in G : ab=ba$$ My question is, can anybody give a proof, or a counter-example (because I don't think it holds) of this theorem? Thanks!
92415
We want to prove that a group, say $G$, of order $pqr$ where $p \gt q \gt r$ has a normal Sylow $p$-subgroup and deduce that it has a normal subgroup of order $pq$.    I know how to show $G$ has a normal Sylow $p$-subgroup or normal Sylow $q$-subgroup, but not the rest. Thanks for your advices.
92507
Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite.  Let $G$ act on left cosets of $H$ by left multiplication.  Let $N$ be the kernel of the action.  Show that $[G:N]$ divides $n!$.  Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$.  So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.  Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$.  Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$...    As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.  Can anyone give some guidance?  Thanks.
88559
Let H be a subgroup in a group $G$. If $G$ is generated by a set $S$, so $G=\langle S\rangle $, prove that $H$ is a normal subgroup in $G$ iff $sHs^{-1}$ $\subseteq H$ for all $s\in S$.    Note with the question: Results of this sort are useful because it is easier to check an algebraic property for a small set of generators than to prove it holds for all elements of G.  Hi, the forward direction is quite straightforward but I will be really grateful if somebody can help me with the reverse direction. Thanks!
85234
How can I prove that, if we have a group G, then subgroup of G generated by all n-th powers of elements from G is normal subgroup of G?