Sorry, I'm new on this subject. I faced a question which let me confused with the concept of a Sylow p-subgroup. The p-subgroups of a group are all contained in the Sylow p-subgroup? In another words, the Sylow p-subgroups are unique?  Thanks
123382
While we know that a unique Sylow p-subgroup is normal, is the converse true?  If there are multiple p-subgroups for the same p are they inherently not normal as a result?
78054
Is the following correct way of showing that there is no simple group of order $pq$ where $p$ and $q$ are distinct primes?  If $|G|=n=pq$ then the only two Sylow subgroups are of order $p$ and $q$.  From Sylow's third theorem we know that $n_p | q$ which means that $n_p=1$ or $n_p=q$.  If $n_p=1$ then we are done (by a corollary of Sylow's theorem)  If $n_p=q$ then we have accounted for $q(p-1)=pq-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.  Is that correct?
115860
I was wondering if there is any classification for normal subgroups of $SU(n)$?  In particular, I think that the answer is no for $n = 2$ by looking at the covering map onto $SO(3)$, but I was curious if there were any results for arbitrary $n$?  I'm actually looking for finite normal subgroups of the unitary group (a lot to ask for, I know).
31401
I tried getting $n_{2}$ and $n_{7}$ , denote the number of Sylow-2-Subgroups and Sylow-7-Subgroups respectively.  I got two cases for $n_{2} = 1 , 7 $ and for  $n_{7} = 1 , 8$ , i noticed that if $n_{2} = 1$ and $n_{7}= 1$ ,then we are done since they are unique subgroups and hence will be normal in $G$.  Next how to proceed with the cases of $n_{2} = 7$ and $n_7 = 8$.  Please help?
119588
In the book I'm study is written:     A normal subgroup of a group need not be characteristic.   And as an exercise I'm supposed to find an example, it also said that is pretty hard to find one. After trying for two days I wasn't able to find one example.   So, I'm asking for an example of a group $G$ with a normal subgroup $H$ that is not a characteristic of $G$.  I will add some context because maybe it will clarify why the book said it is difficult to find an example: the main problem with the exercise is that until it the book had only covered Group Definition, Subgroups, Lagrange's Theorem and Homomorphisms. So I'm supposed to find a example with such lack of advanced tools.
105562
Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$  Question: Deduce that if $G$ is a finite $p$-group, for some prime $p$, the number of subgroups of $G$ that are not normal is divisible by $p$.  Comments:    The subgroups of $G$ that are normal have the property that $g\cdot H=gHg^{-1}=H$ The question is equivalent to showing $p$ divides $\left|G\right|-\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$ $p$-group: $\forall g\in G,|g|=p^k$ for $k\in\mathbb{N}^+$ I don't know where to start with this problem, been thinking about it for a while to no avail, I feel that there are too many definitions for me to consider when finding a solution. My problems I have considered and not been able to answer are: what is the order of $|G|$? I think it must be of order that is divisible by $p$, hence the question becomes show $p$ divides the order of $\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$. How do you compute the order of this set? It seems like quite a standard problem, so I would really appreciate it if I could be directed to more information about whatever problem it is.
22076
I want to prove the following:     If every Sylow subgroup of $G$ is a normal subgroup, then $G$ is isomorphic to the product of its Sylow subgroups.   So far, I have gotten to realize that for every prime factor $p_i$ of $|G|$, there is exactly one Sylow subgroup and that these form a partition of $G$. However, I am having trouble finding the isomorphism between $G$ and the product which I will call $G_1 \times \ldots \times G_r$. My idea was just to use the mapping $(g_1, \ldots, g_r) \mapsto g_1\ldots g_r$, which is a bijection, but I don't see whether it is a homomorphism. In general, not all elements of $G_i$ commute with elements of $G_j$, so one cannot simply change the order of the elements in the product.
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.
89495
Let $G$ be a group and $H$ a subgroup with $[G:H]=n$. Is it true that $x^n\in H$ for all $x\in G$?   Remarks. The answer is positive whenever $H$ is normal, e.g., for $n=2$. In general, by using the normal core of $H$, one can find an $m\ge 1$ such that $x^m\in H$ for all $x\in G$.
52206
If we have a two lattices (partially ordered) - one for subgroups, one for factor groups, and we know order of the group we want to have these subgroup and factor group lattices, is such a group unique up to isomorphism (if exists)? Or is there a counterexample?  If that's true, are sufficient conditions on the order and subgroup lattices to guarantee uniqueness? Another way - what if we now lattice for subgroup and group of automorphism of group; is that group uniquely determined by that information?  Thanks for help. (sorry for English)
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...
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 $G$.
121430
Let $G$ be a finite group and suppose that $\phi$ is an automorphism of $G$ such that $\phi^3$ is the identity automorphism. Suppose further that $\phi(x)=x$ implies that $x=e$. Prove that for every prime $p$ which divides $o(G)$, the $p$-Sylow subgroup is normal in G.  Its an problem from Herstein, prob 19 in 2.12,2nd edition.I need some hint to solve this.Thank you.
117215
By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.   My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?   Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists.
105312
I'm reading Hungerford's algebra and I'm on Nilpotent and solvable groups chapter. Hungerford starts with:     Consider the following conditions on a finite group G:      i)   G is the direct product of its Sylow subgroups      ii)  If m divides |G|, then G has a subgroup of order m      iii) If |G| = mn with (m,n) = 1, then G has a subgroup of order m   This chapter comes after Sylow theorems, where we have seen that if $|G| = p_1^{\alpha_1}p_2^{\alpha_2} \dotsm p_n^{\alpha_n}$ there is a subgroup of order $p_i^k$ for $k \in \{0, 1, ..., \alpha_i\}$, but we can't guarantee more (take for an example $A_4$, it has subgroups of order 3 and 4, but there is no subgroup of order 6). So the question What kind of groups have nice subgroup structure? isn't that unexpected.   What was unexpected?     We shall first define nilpotent and solvable groups in terms of certain       ''normal series''of subgroups. In case of finite groups, nilpotent groups are   characterized by condition i) and solvable ones by condition iii).   Hungerford first gives definition of nilpotent groups in terms of ascending central series and definition of solvable groups in terms of commutator subgroups and then proves equivalences with i) and iii).  Conceptual jump between i), ii), iii) and definitions involving normal series is a little too big for me. I can't really see how someone could start thinking about i), ii), iii) and come out with normal series. I don't have a nice way to think about them and I would really be thankful if someone has a nice conceptual view about them, or several eye-opening exercises.
45770
Find the three 2-Sylow subgroups of $S_3$ and find a 2-Sylow subgroup and a 3-Sylow subgroup of $S_4.$  I just learned Sylow' theorem at the moment and I don't know how to do these problems. I know a p-Sylow subgroup of a group $G$ is a subgroup of order $p^k$, where $p^k$ divides the order of $G$ but $p^{k+1}$ does not. $|S_3| = 6 = 2\dot\ 3$ and $|S_4| = 24 = 2^3\dot\ 3$. How can I solve these types of problems?
5675
I have the following question:      Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$   Then it holds: $g^{n!}\in H$ for all $g\in G$.   Why is this true?  I think, that's not very difficult, but I have no idea at the moment.   Thanks!