Almost all the books of algebra or group theory give following types of applications of Sylow theorems:  A group of order $...$ is not simple. A group of order ... has normal Sylow subgroup.  Even in math.stackexchange, there are many questions with above title and they involve questions which are related to proving above type of statements. Are there any other type of applications of these theorems of Sylow?
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The theorems of Sylow are very well known and almost every mathematician learns in his undergraduate course.  The applications of Sylow theorems given in books are of the kind      "If $|G|=....$ then show that $G$ is not simple/ $G$ is solvable/ ..."   I would like to know if there are some other, interesting, applications of this theorem.
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Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?   says "not much":     There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.   On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:   What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?     Wikipedia then states:     If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$   which is easy to prove, but not very effective. So:   What are some similar results about the order of a product, perhaps with some additional hypotheses?
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Let G be a group of order $729$.   $729 = 3^6$ so by Sylow's Theorem G has a Sylow $3$-subgroup of order $729$. And there are $x$ of them. $r \equiv 1 \pmod 3$ and $r\ |\ 1$. So $x=1$? Is this correct?
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I would like to know whether there are examples where finite group theory can be directly applied to solve real world problems outside of mathematics.  (Sufficiently applied mathematics such as cryptography, coding theory, or statistics still count.)  Let me clarify: I am not interested in applications of elementary group theory which happen to involve finite groups (e.g. cyclic/dihedral/easy groups as molecular symmetries).  I am interested in applications of topics specifically coming from finite group theory as a discipline, like one might see in , , or .  "The Schur multiplier has order 2640, so we should point the laser that way."   "Is this computer system secure?"  "No - Frobenius kernels are nilpotent."  I'm aware of , but many of the applications listed there are inside mathematics or fall in the "applications of easy groups" category.  It is entirely possible that what I'm looking for doesn't exist, and that finite group theory is still an untouchable, pure subject, like number theory in the days of G. H. Hardy.  But perhaps not.  Does anyone know of any applications of the higher level stuff?
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How can we prove that "Every Group of order $216$ contains a normal subgroup of order $27$ or $9$ ?".  MY TRY: $216=2^3.3^3$.So possible order of sylow $3$ subgroup is $1$ or $4$ and similarly for sylow $2$ subgroup is of order $1,3,9\,\text{or}\,27$.Then how to proceed $?$Thank you
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Following the advice given in , I have started to study Group Theory from the very basics. My reference text is Abstract Algebra by Dummit and Foote. While going through the exercises (page 24) I found one problem which required more effort than others:  Assume $G = \{1, a, b, c\}$ is a group of order $4$ with identity $1$. Also assume that $G$ has no elements of order $4$. Use the cancellation laws to show that there is a unique group table for $G$. Deduce that $G$ is abelian.  I proceeded with solution as follows. Since $G$ is of even order there must be at least one element of order $2$ (this statement itself was one of the problems in the exercises and is easy to handle). From the wording of the question it seems that none of the elements $a, b, c$ is singled out with specific properties and hence $a, b, c$ must behave in exactly the same manner. Hence each of them is of order $2$. Thus $a^{2} = b^{2} = c^{2} = 1$.  Next I analyze product $ab$. Clearly it can't be $a$ (as $b$ is non-identity), or $b$ (as $a$ is non-identity) or $1$ (because $ab = 1 = aa$ implies $b = a$). Hence $ab = c$ and similarly $ba = c$. By the similar nature of all elements $a, b, c$ it follows that $bc = cb = a$ and $ca = ac = b$. Thus the operation of the group is defined properly for all elements and clearly it is abelian.  I think my solution is OK, but I am not sure. Note that I have not used the fact that $G$ has no elements of order $4$. Please let me know :  1) if my solution is correct. If not then point out the flaws.  2) if there is any better / shorter solution. If so provide hints and not solution.  The question may sound too easy / simple but I request to treat me like a beginner who has never heard of Group Theory and is reading fresh from the reference text I mentioned. Thanks in advance for your inputs.  EDIT: Thinking further about this problem I wondered what would happen if an element of order $4$ was allowed in $G$. In that case I think that the group has to be isomorphic to the group $G_{1} = \{1, 2, 3, 4\}$ with modulo $5$ multiplication as the group operation. The argument is as follows. Suppose $a$ is of order $4$ and let $b$ be its inverse then $b$ is also of order $4$. Now $c$ has to be its own inverse so that $c^{2} = 1$. Again $a^{2} \neq 1$ (as $ab = 1$), $a^{2} \neq a$ (as $a \neq 1$), $a^{2} \neq b$ (as it would mean $b^{2} = 1$ and thus $b$ would be of order $2$). Hence $a^{2} = c$ and similarly $b^{2} = c$. Next we can see that $ac \neq a$, $ac \neq c$, $ac \neq 1$ (as $ab = 1$) so that $ac = b$. Similarly $ca = b$, $bc = a$, $cb = a$. Also it is clear that this group turns out to be cyclic with both $a$ as well as $b$ as generators. And it also follows that there are the only two ($G$ in original question and $G$ in new variation) groups of order $4$ upto isomorphism.  I hope I have learnt something from the answers given for the original question and this solution is correct. Please let me know if there is any problem with this reasoning.
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In relation to , I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$.  It is easy to see that cyclic subgroups of $D_n$ is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of $D_n$ is normal.  A little bit of Internet search suggests the use of semidirect product $(\mathbb Z/n\mathbb Z) \rtimes (\mathbb Z/2\mathbb Z) \cong D_n$, but I do not know the condition for subgroups of a semidirect product to be normal.  I would be grateful if you could suggest a way to enumerate the normal subgroups of $D_n$ that does not resort to too much of case analysis.
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This is almost a duplicate of the following questions (but, read further):      But, I wasn't able to apply the proof to my case of a group of order 10 and an element of order 5.  In my case, if I try a proof by contradiction, I'm left with the fact that every element in $G$ is of order 2, but if I divide them into subgroups, every subgroup is of order 2, and has one unique element.  I also tried other methods to prove the theorem, but couldn't come up with anything meaningful.  I'm a beginner in the field, and am not familiar with theorems such as Cauchy's or Sylow's theorem (I'm aware of Lagrange's theorem). Therefore, please try to provide a simple answer, thanks!
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Exercise from Herstein:Abstract Algebra.   Please do it without group action as I don't know it. In Herstein there is no mention of Group Action as in the answer given          Show that a group of order $108$ has  a normal subgroup of order $9$ or $27$.      Attempt: $108=2^2\times 3^3$. If $n_2$ denotes the number of Sylow $2$ subgroups then $n_2=1+2k| 27\implies n_2=1,3,9,27.$  If $n_3$ denotes the number of Sylow $3$ subgroups then $n_3=1+3k| 8\implies n_3=1,2,4,8.$  If $n_3=1\implies $ the group has a normal subgroup of order $27$ but how to neglect the other choices.  I am confused totally .Please help.
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If $G$ is a group of order $6$. How do I show that there is some $x\in G$ with $x^2\neq e$?. I was thinking of trying a contra positive approach, by proving that if there is some $x^2=e$ then $G$ is Abelian. But how would I continue after showing that $x^2=e$?
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I'm dealing with the following question: having a group $G$ (of which nothing else is specified, doesn't have to be abelian), prove that $|ab|=|ba|$ for any $a$, $b$. Since $G$ is not abelian I'm not sure what to use here. I was thinking that except the case where $b=a^{-1}$, the order of those would be $lcm(m,n)$ where $m$ and $n$ are the orders of a and b. But looking it up it seems not to be so simple, even if all the info I could find was for abelian groups, so I would like some help with this specific case.
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So I'm having trouble with this problem. I know that the definition of a simple group means that the group has no nontrivial subgroups. I know that this can be proven somehow with the help of the converse of Lagrange's Theorem for Abelian groups: If G is abelian of order n, and d is a divisor of n, then G has a subgroup of order d.   My attempt: (=>)Assume that G is a finite abelian group and is simple, then G has no nontrivial normal  subgroups.  (Now I don't know how to show that this implies that G has order p, where p is prime.  (&lt;=)Assume that G is a finite abelian group with order p, where p is a prime. (Since the order of p is prime then what does this mean?)  Edit: Can someone check my new attempt at the proof?   (=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and  is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order.   (&lt;=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups.
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I don't know about the Sylow Theorems.  But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.  If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$.   But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.  I am asking for help completing this argument.  Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).
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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 $|G|=8p$.
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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.
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I'm looking for a simple proof that up to isomorphism every group of order 2p (p prime) is either $\mathbb{Z}_{2p}$ or $D_{p}$ (The Dihedral group of order 2p).  I should note that by simple I mean short and elegant and not necessarily elementary. So feel free to use tools like Sylow Theorems, Cauchy Theorem and similar stuff.  Thanks a lot!
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Question is to :  Find the number of Normal subgroups of a nonabelian group $G$ of order $21$ other than $\{e\}$ and $G$.  What I have done so far is :  As $|G|=21=3\cdot7$  we have :  No. of sylow $3$ subgroups $1+3k$ dividing $7$ leaving out possibilities $1$ or $7$  No. of sylow $7$ subgroups $1+7k$ dividing $3$ leaving out only possibilities $1$.  So, we have a unique sylow $7$ subgroup and so it is normal.  I remember somehow that any normal group should come from a normal sylow subgroup or something like that.  S0, I prefer to conclude there is only one Normal subgroup for a non abelian group of order $21$.  Please let me know if this is true and please help me to fill that gap :      I remember somehow that any normal group should come from a normal sylow subgroup or something like that.   Thank you.