$|G| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. I know this might have been asked and answered before. I just wanted someone to tell me if my argument is OK:  Let $|G| = pqr$, and assume $p &lt; q &lt; r$. We have at least one Sylow-$q$ subgroup (call it $Q$) and at least another Sylow-$r$ subgroup (call it $R$). Consider the subgroup $K = $ $&lt;Q, R&gt;$. Now $K$ is not the whole group because there is no element of order $p$ in there. Further, $|K|$ must equal $qr$ because nothing smaller is possible by Lagrange's theorem.  Therefore [G : K] = p, which is the smallest prime dividing G. Hence by a standard theorem (for instance, see corollary 4.5, p. 44, Isaacs' Algebra), K must be normal. Hence $G$ is not simple.  Thanks!  WH
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
113638
Let $G$ be a finite group. Let $H \trianglelefteq G$, with $\vert H \vert = p$, a prime, where $p$ is the smallest prime dividing $\vert G \vert$. Prove that $H \leq Z(G)$. (Hint: If $a \in H$, by normality, its conjugacy class lies inside $H$.)   My approach so far:  Let $a \in H$. Then $C(a) = \{gag^{-1} : g \in G\} \subseteq H$. Now, since $\vert H \vert$ is prime, it follows that $H$ is cyclic and moreover $H$ is abelian, hence $C_H(a) = H$ for all $a \in H$ and equivalently $[H: C_H(a)] = 1$ for all $a \in H$.  Try to show $\vert C(a) \vert = [G:C_G(a)] = [H:C_H(a)] = 1$.  So, since $\vert C(a) \vert = 1$ for all $a \in H$, we have $C(a) = \{a\}$ for all $a \in H$ and equivalently $a \in Z(G)$ for all $a \in H$, so we can say $H \subseteq Z(G)$ and moreover $H \leq Z(G)$.  My main question is, how to show $[G:C_G(a)] = [H:C_H(a)]$. But also, I haven't used the fact that $p$ is the smallest prime dividing $\vert G \vert$, so is my reasoning wrong anywhere?
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.
116528
Proposition: Given a group $G$ with $|G|=24$ such that $\nexists g\in G$ with $|g|=6$, then $G\cong S_4$.    I understand methods you can employ to deduce the number of Sylow $p$-groups in $G$ by counting elements or reasoning about permutation representations. But how can we construct an isomorphism to $S_4$ given that $n_{2-\text{Sylow}}=3$ and $n_{3-\text{Sylow}}=4$? Or otherwise, rule out all other possible cases (perhaps reasoning with Cayley's Theorem)? I am also interested in finding a proof that is perhaps less direct, but more elegant, in particular one that may involve the irreducible representations of $S_4$. Any insight appreciated.
99853
I am trying to solve the above question, as an application of Sylow's theorem.  Let $P$ be the p-Sylow subgroup. Then $n_p | (p+1)$ and $n_p \equiv 1 \pmod{p}$. If $n_p =1$, $P$ is normal and we are done, else $n_p = p+1$. Now,   \begin{equation} 1+n_p(p-1) = 1 + (p+1)(p-1) = p^2, \end{equation} is the total number of elements in the p-Sylow subgroups. So if $n_p = p+1$, that means the number of elements not in any p-Sylow subgroup is $|G|-p^2=p$. If these $p$ elements and the identity form a subgroup then its a subgroup of the smallest prime index, so we are done.   But how do I show that all the elements not in the p-Sylow subgroups form a subgroup, i.e. the subgroup generated by these elements has trivial intersection with the $p-$Sylow subgroups?
86070
I have this following question from my class note on Sylow Theorem:     Show that a group of order 30 can not be simple.   For that I know the followings: (1) A simple group is one that does not have non-trivial normal group,  (2) Group $G$ is p-group if there exists an integer $e$ such that $|G| = p^e,$  (3) A $p$-subgroup $H$ of $G$ is called Sylow p-subgroup of $G$ if $p \nmid |G/H|,$ and finally  (4) A subgroup $H$ of $G$ is a Sylow p-subgroup if there exist integers $e$ and $m$ such that $|G| = p^em,$ and $(e, m) = 1,$ and also $|H| = p^e,$  but unfortunately I don't know how to weave them into a solution, any help would therefore be very much appreciated. Thank you for your time and help.  PS. If you have choices of elegant versus dummy down-to-earth solutions, do please give me the latter. I know it would be tedious for you but you have a slowpoke turtle down here. Thank again for your patience.  POST SCRIPT : ~~~~~~~~~~~~~~~~~~~~~~~  I came across a very simple solution by John Beachy , consisting of only 2 lines, which I adapted to fit group order of 30:  EDIT (j.p.) Unfortunately the linked solution cannot be easily adapted to this case here. I kept the text as far it was correct.  (1) Recall Sylow's Theorem: If $|G| = p^em; p, e, m \in \mathbb Z^+; (p, m) = 1;$ the number of $Syl_p (G) = s$, then $s \equiv 1\pmod p$ and $s \mid m.$  (2) For $|G| = 5 \cdot 2 \cdot 3$, suppose that $p = 5, e = 1$ and $m = 6.$ Therefore $s \equiv 1\pmod 5 $ and $s \mid 6,$ thus $s = \{1, 6 \}$. The 5-Sylow subgroup is normal if and only if $s=1$, and in this case $G$ is not simple. (3) Now look at .  Hope this will be useful for some of you and thank you very much for those of you who have helped me.
3395
I'm having trouble with an exercise in "Introduce to the theory of group" book. This is problem:  Let $M$ be a maximal subgroup of $G$. Prove that if $M$ is a normal subgroup of $G$, then $[G: M]$ is finite and equal to a prime.
14728
Suppose there are $2$ permutations given by $p$ and $q$. I need to check whether $p$ belongs to the group generated by $q$, and if so, it's representation in a power of $q$. In other words, I've been given $p$ and $q$, I need to check whether $\exists i \in \{ 0,...,|q|-1 \} \text{ such that } p = q^i$ I know that if $m = \gcd(i, |q|)$, then $|q^i| = \frac{|q|}{m}$. Now, say that such an $i$ exists. Then, $$ |p| = \frac{|q|}{\gcd(i, |q|)} \Rightarrow \gcd(i, |q|) = \frac{|q|}{|p|} $$ Thus, a necessary condition for such an $i$ to exist is: $|p|$ divides $|q|$ Say that this holds, and $\frac{|q|}{|p|} = r$, where r is an integer. Then we just have to solve $\gcd(i, |q|) = r$ for $ i \in \{ 0,...,|q|-1 \}$. Can this be done? If not manually, then by some efficient algorithm? (Note: Just looping over all values of $i$ is not efficient as $i$ may be very large) Edit: Another way I can think of solving this is by writing both $p$ and $q$ as product of disjoint cycles: $$ p = c_1c_2\ldots c_k \text{ and } q = d_1d_2\ldots d_l $$ And thus, as disjoint cycles commute: $$ p^i = c_1^ic_2^i\ldots c_k^i $$ How to proceed after this?
22464
I would be very thankful if someone could give me a hint with proving this. It's a very common exercise in abstract algebra textbooks.     If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $K$ contained in $H$ whose index in $G$ is finite and divides $n!$.   I found the proof on  at some point (although the proof appears to be no longer there), but I got lost in one of the details.
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 $|G|=8p$.
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.
41977
Let $G$ be a group of order $56$. (We do NOT assume the Sylow-$7$ subgroup to be normal.) Then either the Sylow-$2$ subgroup is normal or the Sylow-$7$ subgroup is normal. How to prove?  My idea: consider the case $n_2=7,n_7=8$. Then the number of elements in all Sylow $2$ -subgroups and Sylow $7$-subgroups is at most exactly $(7-1)*8+(2-1)*7+1=56$. This cannot give a contradiction. We need to prove that $G$ is simple. Then $|G||(8!/2)$. To prove $G$ simple, we need to prove there does not exist normal subgroups of order $2$ and order $4$. But I do not know how to prove.
19200
Question. Let $G$ be a finite group and $p$ be the smallest prime dividing $|G|$.  Let $x$ be n element of order $p$ in $G$. Assume that there exists an element $h\in G$ such that $hxh^{-1}=x^{10}$.  Show that $p=3$.    What I have done so far is use the fact that $o(hxh^{-1})=o(x)=p$ and $o(x^{10})=p/\gcd(p,10)$.  By hypothesis we have $p=p/\gcd(p,10)$, giving $\gcd(p,10)=1$.  Here I am stuck.  I have not been able to make use of the fact that $p$ is the smallest prime divisor of $|G|$.  Can somebody help?
48145
The question states:     Let $g$ by a primitive root of the odd prime $p$. Show that $-g$ is a primitive root , or not, according as $p \equiv 1 \pmod 4$ or not.   For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated.  Thanks
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
115391
Suppose that G is a group and N &lt; G a normal subgroup. Assume that there is no normal subgroup M &lt; G with N &lt; M. Prove that G/N is simple.  This problem is really throwing me for a loop. The concept of quotient groups and how they relate to simple subgroups is confusing me, and so I am not really sure how to begin. If a simple subgroup's only normal subgroup is itself (and the trivial group), then G/N being simple would be proving that when multiplying the entire subgroup N by an element g for all g in G, that the collection of those sets cannot be broken into smaller normal subgroups?
42433
I'm trying to figure out this problem:  Problem: Let $G$ be a group of order $p^k q$ with $p$ and $q$ distinct primes and $k \in \mathbb{N} \setminus \left\{0\right\}$.   a) Prove: $G$ has a subgroup of order $p^k$, or $p$ divides $|Z(G)|$. (Here $Z(G)$ means the center of $G$).   b) Prove: if $p$ divides $| Z(G)|$, there there exists an $x \in Z(G)$ of order $p$.   Attempt: For a) I tried proceeding by induction.  Base step: Assume $k=1$. Then $|G| = pq$ with $p \neq q$. Case 1: if $G$ is commutative, then $Z(G) = G$ and then $p$ divides $|Z(G)|$.   Case 2: Assume $G$ is not commutative. Then the center of $G$ is trivial, i.e. $Z(G) = \left\{e_G\right\}$ (this is a theorem of group theory). So $p$ does not divide $|Z(G)|$. Then we must show that $G$ has a subgroup of order $p$. Let $H$ be a non-trivial subgroup of $G$, that is $H \neq G$ and $H \neq \left\{e\right\}$. Then by Lagrange, $|H|$ divides $pq$. So we must have either that $|H| = p$ or $|H| = q$ (is this correct? Since $H$ is non-trivial). If $|H| = p$ then we are done. If $| H | = q$, then $|G / H| = p$ and so $G/H$ is cyclic. I want to derive a contradiction from this, and so show that $H$ cannot have order $q$, but I don't know how.  Induction step: Assume the claim holds for $|G| = p^{k-1} q$. We must then show that it also holds for $|G| = p^k q$.  I want to use the class equation. We have that $$ p^k q = |Z(G)| + \sum_{i=m+1}^n |Cl(a_i)| $$ where $|Cl(a_1)| = \ldots = |Cl(a_m)| = 1$ and $|Cl(a_i)| &gt; 1$ for $i &gt; m $. Since $p$ is a divisor of the LHS, it must also divide the linear combination. But how can I conclude from this that $p$ divides $| Z(G)|$ ? I don't know how to use the induction hypothesis.   b) Assume $p$ divides $|Z(G)|$. I first assume that $G$ is abelian. Then by structure theorem for finite abelian groups, we know that $$ G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \ldots \oplus \mathbb{Z}_{p_r^{k_r}}. $$ This means that $$ p^k q = p_1^{k_1} \ldots p_r^{k_r}. $$ So $p | (p_1^{k_1} \ldots p_r^{k_r})$ and so since $p$ is prime, $p$ must divide at least one factor. Assume that $p$ divides $p_i^{k_i}$. Since $p_i^{k_i} = p_i \ldots p_i$ with $k_i$-amount of factors, we must again have that $p$ divides $p_i$. So $p = p_i$.   Now, since every $\mathbb{Z}_{p_i^{k_i}} = \mathbb{Z}_p^{k_i}$ is abelian, we have that the center equals the whole group. Let $x_i$ be an element in this center. Then $order(x_i) | p^{k_i}$ and so $order(x_i) | p$. Hence the order of $x_i$ equals $p$ in $\mathbb{Z}_p^{k_i}$. Since $x_i$ is also in $Z(G)$, this concludes the proof?   And what about the case that $G$ is not abelian?  Help with this problem is appreciated!