Let $G$ be a group of order $399$. Then by Sylow, it must have a normal subgroup of order $19$, denoted by $H$. Let $N, K$ be groups of order $7$ and $3$. Then $HN$ and $HK$ are groups of order $133$ and $57$. What is the possible order of their normalizer?  The normalizer can only be either themselves or the whole group by Lagrang
120637
Here's a question I've been asked;  Given distinct primes $p,q,r$, show that any group $G$ of order $pqr$ is not simple.   So far, my idea has been to individually check each possible proper subgroup, since by Lagrange there are only $$\dbinom{3}{2} + \dbinom{3}{1} + 1$$ of them, and show that at least one of them other than the trivial group is normal. Surely there is a smarter way, since generalizing this approach wouldn't work so well if I wanted to show similar results for groups who's order is a product of $n$ distinct primes.  Is there some other approach? Thanks for any insights
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.
35999
I wanted to share with you my resolution of this exercise.     How many different groups of order $15$ there are?   My resolution:     We're looking for groups such that $|G|=15=3\cdot 5$. Then: $G$ has an unique Sylow $3$-subgroup of order $3$ ($P_3$) and an unique Sylow $5$-subgroup of order $5$ ($P_5$).      Furthermore, $$|P_3 \cap P_5|\  \Large\mid\ \normalsize|P_3|,\ |P_5|$$   So $$|P_3 \cap P_5|=1\implies P_3 \cap P_5=\{e\}.$$      Then:$$|P_3\cdot P_5|=\displaystyle\frac{|P_3|\cdot |P_5|}{|P_3 \cap P_5|}=|P_3|\cdot |P_5|=3\cdot 5=|G|,$$      and $$G=P_3\cdot P_5 \simeq P_3 \times P_5 \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}.$$   It's everything correct and well justified? There's some things that I didn't say, for example that a subgroup a prime $p$ as order is cyclic, and thus isomoprhic to $\mathbb{Z}/p\mathbb{Z}$. Maybe I did something wrong. I would appreciate someone to look and criticize my exercise.  Thank you.
35550
Show that any group of order 200 is not a simple group.   I have started with the sylow 5-subgroup, but the sylow 2 subgroups i found that they are 25 subgroup but i couldnt proceed in the proof any more! Please help.
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 $G$.
54470
Let $G$ be a simple group and there exists a subgroup $H$ of index $n \in \mathbb{N}_{\ge 3}$. Show that: $|G|$ divides $n!/2.$ I saw a similar question .   There was written: "The proof is really simple. Let $X$ be the set of left cosets of $H$. Consider $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$. Then $\phi$ is a homomorphism. Consider now $K=\ker \phi$. Then $K$ is a normal subgroup of $G$ contained in $H$. Finally, $G/K$ is isomorphic to a subgroup of $\text{Sym}(X)$, which has order $n!$, where $n=[G:H]$. Thus, $[G:K]$ is finite and divides $n!$."  In fact, I don't understand the last sentence. "Thus, $[G:K]$ is finite and divides $n!$." Also did I understand it right that $ker(\phi)=H$?  Can you also give me an advice how to proof it for $n \ge 3$. (maybe induction)  Thank you for taking your time.
44880
Consider the following question:     Show that a group $G$ of order $8p$ is solvable, for any prime $p$.   I am kind of stuck, but here are my first attempts:  I chose the series of subroups $G&gt;H_8&gt;H_4&gt;H_2&gt;H_1=\{e\}$ where $H_k$ has order $k$. All of these subgroups exist due to Sylow's Theorems. We have the quotients $H_8/H_4\cong H_4/H_2 \cong H_2/H_1 \cong \mathbb Z/2\mathbb Z$, so $G&gt;H_8\rhd H_4\rhd H_2\rhd H_1=\{e\}$. Only the factor $G/H_8 \cong \mathbb Z/p\mathbb Z$ is causing me a headache, because although its of prime order (and hence abelian) I don't know whether $H_8$ is normal in $G$ (unless $p=2$).  If $p\ne 2$, then the number $k$ of Sylow $2$-subgroups is $1$ mod $2$. Since $k$ divides $|G|$ we might have $k=1$ or $k=p$. If we had $k=1$ then $H_8$ would be the only Sylow $2$-subgroup (and hence normal in $G$). But how could we show this?  Or is there even an easier way to approach this problem?  $\ $  edit: I am going to try a different approach: The case $p=2$ is clear. When $p=3$ or $p=7$ the group has order 24 or 56 and since the smallest simple non-abelian group has order 60, the group must also be solvable in these exceptional cases (as Mariano pointed out).  In any other case we get a Sylow $p$-subgroup $H_p$ of order $p$, which is is normal in $G$. The quotient $G/H_p$ has order $8=2^3$; and prime power order implies solvable. So $G/H_p$ is solvable and $H_p$ is also solvable. And since the composition factors of $G$ are those of $H_p$ together with those of $G/H_p$, we conclude that $G$ is solvable iff $H_p$ and $G/H_p$ are solvable, which is the case. Hence $G$ is solvable if $|G|=8p$.
78268
I want to show that there are no simple groups of order $p^{k}(p+1)$ where $k&gt;0$ and $p$ is a prime number.  So suppose there is such a group. Then if we let $n_{p}$ denote the number of $p$-Sylow subgroups of $G$ we have that $n_{p}=p+1$. Now by letting $G$ act on $Sylow_{P}(G)$ by conjugation we obtain a group homomorphism $G \rightarrow S_{p+1}$. Since $G$ is simple then either $ker(f)$ is trivial or all $G$. Now here's my question: assume $ker(f)=G$ this would imply then that $G$ has a unique $p$-Sylow subgroup no? but then such subgroup is normal which contradicts the fact that $G$ is simple. So the map in fact is injective but then $|G|$ divides $(p+1)!$ which cannot be.  Basically my question is if my argument is correct, namely thta if $ker(f)=g$ implies the existence of a unique $p$-Sylow subgroup which implies such subgroup is normal in $G$ which cannot be. In case this is wrong, how do you argue that $ker(f)$ cannot be all $G$?  Thanks
74874
If $N$ is a normal subgroup of $G$ and $H$ a characteristic subgroup of $N$ (i.e., $\sigma(H)=H$ for all $\sigma \in \operatorname{Aut}(N)$). Show that $H$ is normal in $G$.     Does one needs to show that $\sigma(gHg^{-1})=gHg^{-1}$ for every g in G. Would that imply H is normal? is there a alternative way?
54677
How can I show that if some group $G$ has only one subgroup $K$ of order $n$, then $K$ is a normal subgroup? Would that mean that it only has one subgroup total? If so then I guess that makes sense.
43160
solution :-  $21= 3 \times 7$  there is only one Sylow $3$ and Sylow $7$ subgroup  so, Sylow $3$ and Sylow $7$ subgroup are normal in group $G$  so $G$ is cyclic group of order $21$.  Am I right ?  somebody told me that group of order $21$ is not cyclic.  he gave me this .  If group of order 21 is not cyclic, then can we understand it by Sylow method ?
35429
I am struggling with the following problem:     Find a group $G$ such that whenever $m, n, k \geq 2$ are natural numbers, then there exist $a, b \in G$ such that the order of $a$ is $m$, order of $b$ is $n$, and order of $ab$ is $k$.   The group $G$ necessarily has to be non-commutative and infinite.
39200
$G$ has no subgroup with index 2. How can I show that all subgroups with index 3 are normal? (The index of the subgroup is the order of the quotient set.)
48825
I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.    We have that  $$ (gh)^{mn}=g^{mn}h^{mn}=1 $$  using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|&lt;mn$ and arriving at a contradiction, but it did not take me anywhere.
3988
Generalizing the case $p=2$ we would like to know if the statement below is true.  Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal.
47472
Prove that if $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m &lt;n.$   Since $G$ is of prime-power order I know $|Z(G)| \ne e$ so there is an $a\in Z(G)$ with order $p$ such that $p \mid |Z(G)|$. Now, the subgroup generated by is normal since it's a subgroup of the center. How can I get this normal subgroup to be less than $n$?
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.