if i have a group G, and a normal cyclic subgroup H, any subgroup of H is normal to G? It is to H, but i don't know to G. Can you give me some help with this?
59995
Exercise 11, page 45 from Hungerford's book Algebra.     If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every   subgroup of $H$ is normal in $G$.   I am trying to show that $a^{-1}Ka\subset K$, but I got stuck. What I am supposed to do now?  Thanks for your kindly help.
115446
Find all groups of order $6$.   If there is element of order $6$ then group is cyclic. If not, I came to conclusion that there should be an element of group $G$ of order $3$ (otherwise the order of group is $2^{n},$ for some $n \in \mathbb{N} $). Let that element be $a$. Now, let $ b \in G \setminus\langle a\rangle$. It is easy to show that all elements $ e, a, a^{2}, b, ab, a^{2}b $ are different. Now I am stuck. I am sure that this isn't enough information about group $G$ but since I am a beginner in this scope I don't know what else should I write. Any hint helps!
8067
"Let $G$ be a group, and suppose $G=H \cup K$, where $H$ and $K$ are subgroups.  Show that either $H=G$ or $K=G$."  Let $h \in H$ and $k \in K$.  Then $hk \in H$ or $hk \in K$ (since every element of $G$ is in either $H$ or $K$).  If $hk=h'$ for some $h' \in H$, then $k=h^{-1}h'$, so $k \in H$.  If $hk=k'$ for some $k' \in K$, then $h=k'k^{-1}$ so that $h \in K$.  If for all $h \in H$ we have $h \in K$, or if for all $k \in K$ we have $k \in H$, then $H \subseteq K$ or $K \subseteq H$.  Then since $G=H \cup K$, we must have either $H=G$ or $K=G$.  I'm not sure if the first paragraph of my 'proof' implies the second.  I've shown that for arbitrary $h \in H$, $h \in H$ and possibly $h \in K$, and similar for $k \in K$.  I don't know how to wrap it up (or perhaps this route won't lead anywhere at all).  If this way won't work, I'd just like a hint on a new direction to take.  Thanks.
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!
92045
Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $4$. Prove that $G$ has a subgroup of index $2$ (hence normal) and if $G/H$ is not cyclic then $G$ is equal to the union of three proper subgroups.   I understand that $G/H$ is isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2$. Let $K^*$ be a subgroup of $G/H$ of order $2$. Then $K^*\cong K/H$ for some $K$ such that $H$ is normal in $K$ and hence $[G:K]=[G/H:K/H]=4/2=2$.  If $G/H$ is not cyclic then $G/H\cong \mathbb{Z}_2\times \mathbb{Z}_2$. I don't know how to prove $G$ is the union of three subgroups. Could it be true that as $G/H$ is the union of three proper subgroups $K_1^*,K_2^*,K_3^*$ then the corresponding groups $K_1,K_2,K_3$ have union equal to $G$? Would anyone give me a hint to prove this?  Thank you.
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 $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
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Let $G$ be a group and $H$ be a normal subgroup of index $p$ ( a prime ) ; suppose $K$ is a subgroup of $G$ not contained in $H$ , then is it true that $G=HK$ ? ( I know that the fact is true if $p=2$ , but I don't know for the general case . Please help . Thanks in advance )
83231
Let $A$ be a normal subgroup of a group $G$. Let $B$ be a subgroup of   $A$ with the property that whenever $B_1 \leq A$ is conjugate to $B$   in $G$ then it is conjugate to $B$ in $A$. Prove that $G=AN_G(B)$   Help me some hints.
101715
Suppose that $ N $ is a normal subgroup of a finite group $ G $. If $ |N| = 5 $ and $ |G| $ is odd, why is $ N $ contained in $ Z(G) $, the center of $ G $?  I know how to do this when $ |N| = 2 $ and $ |G| $ is even, but am not sure what to do with this one. Sylow's theorems maybe?  Thank you guys!
52932
Suppose a cyclic group has exactly three subgroups: $G$ itself, $\{e\}$, and a subgroup of order $7$. What is $|G|$? What can you say if $7$ is replaced with $p$ where $p$ is a prime?   Well, I see a contradiction: the order should be $7$, but that is only possible if there are only two subgroups. Isn't it impossible to have three subgroups that fit this description?  If G is cyclic of order n, then $\frac{n}{k} = 7$. But if there are only three subgroups, and one is of order 1, then 7 is the only factor of n, and $n = 7$. But then there are only two subgroups.  Is this like a trick question?  edit: nevermind. The order is $7^2$ right?
114819
If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$ ?    This means, $gHg^{-1}$ is a proper subgroup of $H$. We know that $H \cong gHg^{-1}$, so if $H$ is finite then we have a contradiction since the isomorphism between the two subgroups implies that they have the same order so $gHg^{-1}$ can't be proper subgroup of $H$.  So, what if $H$ is infinite is there an example for such $G , H , g$ ?   Edit: I suppose that $H$ has a subgroup $N$ such that $N$ is a normal subgroup of $G$.
111109
Let $G$ be a group. Prove or disprove that $H =\{g^2 | g \in G\}$ is a subgroup of $G$.   I tried testing the permutations of $A_4$, however squaring each cycle yielded a cycle in $A_4$ so I'm lacking a counter-example (if there is one). In a nutshell I'm looking for a subgroup such that when you square the permutation cycle, it yields a cycle not in that subgroup.  Or, I could be way off base and figure out that there isn't a counter-example and I need to prove that indeed $H$ is a subgroup of $G$.
18609
I'm being asked to find 2 subgroups of $S_3$, one of which is normal and one that isn't normal. I guess, to find the non normal subgroup is easier. I would do this by trial and error, but since the group is $S_3$ and easily visualizable, I guess that there should be a geometrical property that makes it easy to find a normal subgroup. Any of you guys know one? Or should I try this by brute force?
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
37197
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.    I have tried to using the definition but failed. Can someone help me please.
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".
76841
Now as $x$ is of order $4$ and it generates $H=\langle x\rangle$ a cyclic subgroup of $G$ of order $4$, hence index of $H$ in $G$ is $2$.  So $H=\langle x\rangle$ is normal in $G$.  Now $gH=Hg $, for all $g$ in $G$.  => $g(x^2)=(x^2)g\:$ (since  $x^2$ is an element of $H$), for all $g$ in $G$.  Hence $x^2 \in Z(G)$.  This was my approach. (I think that I am wrong in here, $g(x^2)=(x^2)g$) Please, correct me wherever necessary. And if you could provide a better solution, please do.