I need clarification on my understanding of Sylow Theorems.   Can I say that a finite p-group will have a subgroup for each prime power? If the above is valid, can I say then that p-groups satisfy the converse of Lagrange Theorem?   I'm not sure of that statement. Can someone show me a counterexample of it if not? Thanks
15414
It is known that the converse of  isn't true in general. More precisely it is known that the following proposition:  If $G$ is a finite group of order $n$ and $m\mid n$ then there exists a subgroup $H$ of $G$ such that $\operatorname{order}(H)=m$.  isn't true for all finite groups $G$. My questions are:   For which groups $G$ does the converse of Lagrange's Theorem (as stated above) hold? More precisely, if $G$ is a group for which the converse of Lagrange's Theorem as I mentioned above holds then what properties must $G$ satisfy?  If there is no complete classification of such $G$s then can someone give me references to works by other mathematicians where they try to give at least a partial classification of these $G$s?    Please note that I am not interested in knowing a complete classification of the groups for which a partial converse holds ( does the job in some sense). I want to know a complete classification of the groups for which the converse of Lagrange's Theorem as I mentioned above holds.
79390
I have this written down in my notes, but cannot seem to find verification of it. How can I see this fact?   I believe there is a theorem, something along the lines of if a group G is isomorphic to H then, H is a group. Can anybody verify this or point me to the direction I am looking for? Thanks.
32431
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 )
105413
It seems that there are many ways to prove the Sylow theorems. I'd like to see a collection of them. Please write down or share links to any you know.
42069
I'm trying to prove that the converse of Lagrange's theorem is true for finite abelian groups (i.e. "given an abelian group $G$ of order $m$, for all positive divisors $n$ of $m$, $G$ has a subgroup of order $n$"). This is an exercise from a book, and it is in the section on finite abelian groups, so I know I have to use the fundamental theorem of finite abelian groups. I have come up with a proof, but it seems a bit messy, and I'm not entirely sure if it's correct. It is given below.     Let the order of $G$ be $m$ = $p_1^{\alpha_1} \ldots p_k^{\alpha_k}$. It is known that $G$ is a direct product of $p$-groups, say:      $$G = G_1 \times \ldots \times G_k$$      where each $G_i$ is a $p_i$-group. By the fundamental theorem of finite abelian groups, each $G_i$ is isomorphic to a direct product of cyclic groups of the form      $$\mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}},$$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$. Now if $n$ divides $m$, then we must have      $$n = p_1^{\gamma_1} \ldots p_k^{\gamma_k}$$      for some $\gamma_1, \ldots, \gamma_k$ with $0 \leq \gamma_i \leq \alpha_i$.      Claim: Each $G_i$ has a subgroup of order $p_i^{\gamma_i}$      Proof: As above, we have that       $$ G_i \cong \mathbb{Z}_{{p_i}^{\beta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\beta_l}} $$      where $\beta_1, \ldots, \beta_l$ are positive integers such that $\sum_{j=1}^l \beta_j = \alpha_i$.      Now since $0 \leq \gamma_i \leq \alpha_i$, we can find   $l$ numbers $\delta_1, \ldots , \delta_l$ such that $\gamma_i = \sum_{j=1}^l \delta_j$, and $0 \leq \delta_j \leq \beta_j$. (This choice of numbers is not necessarily unique).      Then $p_i^{\delta_j} | p_i^{\beta_j}$ for each $j = 1, \ldots , l$. Hence, for each factor $\mathbb{Z}_{{p_i}^{\beta_j}}$, there exists a subgroup of order $p_i^{\delta_j}$, namely $\mathbb{Z}_{{p_i}^{\delta_j}}$ (using the fact that the converse of Lagrange's theorem is true for finite cyclic groups). Taking the direct product of each of these subgroups, we get a new subgroup $G_i'$ of $G_i$:      $$G_i' \cong \mathbb{Z}_{{p_i}^{\delta_1}} \times \ldots \times \mathbb{Z}_{{p_i}^{\delta_l}}$$      The order of this subgroup is $p_i^{\delta_1} \times \ldots \times p_i^{\delta_l} = p_i^{\delta_1 + \ldots + \delta_l} = p_i^{\gamma_i} $. So we have found a subgroup of $G_i$ of order $p_i^{\gamma_i}$, as required.      So each factor $G_i$ in the product $G = G_1 \times \ldots \times G_k$ has a subgroup $G_i'$ of order $p_i^{\gamma_i}$.       Therefore, $G$ has a subgroup   $$G_1' \times G_2' \times \ldots \times G_k'$$       of order $p_1^{\gamma_i}...p_k^{\gamma_k} = n$, which completes the proof.   I have two questions about this: firstly, does this proof seem to work? Secondly, is there a way to make the proof more concise (e.g. a way to prove the statement without using all these indices)?
43458
I think I remember seeing a theorem about this in my Abstract Algebra class, but I cannot seem to find the reference for it anymore.   Let $G$ be a group and let $H$ and $K$ be non trivial subgroups of $G$. Are there any sufficient conditions on $H, K$ and $G$ such that $G = HK$? (In other words $\forall g \in G$, there exist $h \in H$ and $k \in K$ such that $g = hk$)
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 $n$?
95937
This is not for homework. (I am a grader for a class.)  The case in which $G$ is finite is trivial. (That is, use a corollary to Lagrange's Theorem, and set $[G:H] = \dfrac{|G|}{|H|}$, and similarly for $[H:K]$.) How do you prove this for when $G$ is infinite?   The proof of this statement that I know well tries to show that $[G:K]$ is finite iff $[G:H]$ and $[H:K]$ are finite, but I am not a fan of that proof. For pedagogical purposes, I'm looking for a better proof. (The course uses right cosets by convention.)  The course has just started covering normal subgroups, so things like the Isomorphism Theorems are not known yet.
22984
Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be written as the direct sum $G=\langle g \rangle \oplus H$ for some $H \leq G$ (subgroup of $G$). For a proof see  for example (page 2).  My question: Do we need that $G$ is a $p$-group or does it also work for arbitrary finite abelian groups?   I think it is wrong for general groups because I looked around quite a bit and always only found the above theorem, but I could not find a counter-example.
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.
49739
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!
94665
I have given that, let $G$ be a group of order $455$. Then I have to show that $G$ is cyclic. Then by using Sylow-theorems it can be solved. But my question is, is there any method to see quickly that a group of order $pqr$ with a relation between them, where all of $p$, $q$, $r$ are distinct primes, is cyclic. Just like a group of order $pq$ is cyclic if $q&gt;p$ and $p$ does not divide $q-1$. For example, any group of order $15$ is cyclic just from the above argument. Here $455=5.7.13$. So is there any simple way to say $G$ is cyclic?
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.
1204
I am trying to solve this problem:  Let $G$ be a finite and simple group, and let $p$ be a prime number such that $p$ divides $|G|$. If $n_p(G)=n$ for $n&gt;1$ (n_p denotes the number of p-Sylows) then $G$ is isomorphic to a subgroup of $\mathbb A_n$.  What I thought of was the following:  I consider the action of $G$ on $X$={H subgroup of G, H p-Sylow} by conjugation. Then there is a morphism $$\psi:G \to S(X)$$ $$g \to gSg^{-1} \space \forall S \in X,$$  where $S(X)$ is the set of all bijective functions $f:X \to X$. First of all, I would like to affirm $S(X) \cong \mathbb S_n$, but I don't know if this is true. Second, note that  $Ker(\psi) \neq G$ because if not $n_p=1$ and $Ker(\psi)$ can't be a proper subgroup because if that was the case $Ker(\psi) \lhd G$, then we have $Ker(\psi)=\{e\}$.  By the first isomorphism theorem, $G \cong G/Ker(\psi) \cong Im(\psi)$. Now, if I could show that $Im(\psi)$ is isomorphic to a subgroup of $\mathbb A_n$, then I would be done.  I would appreciate any hints or suggestions.
111586
Let $G$ be a finite group, $p$ a prime number that divides $|G|$ and $O_p(G)=\bigcap_{P \in Syl_p(G)}P$. Prove that  1) $O_p(G) \lhd G$  2) $O_p(G)$ is maximal among the normal $p$-subgroups of $G$.  3) $O_p(G/O_p(G))=1$  I think I could do 2) using the following proposition which I will just enunciate:  Proposition  Let $P$ be a $p$-Sylow and $H \subset G$ a $p$-subgroup. Then there is $x \in G$ such that $xHx^{-1} \subset P$.  Solution to 2):  Suppose $O_p(G) \subset H$ for some $H \lhd G$ p-subgroup. By the proposition we have that for each $P$ p-Sylow group, there is $x: H=xHx^{-1} \subset P$. But then $H \in \bigcap_{P \in Syl_p(G)} P=O_p(G)$, so $O_p(G)=H$, it follows $O_p(G)$ is maximal.  I am pretty lost in 1) and 3), I would appreciate suggestions or hints rather than complete answers. Thanks in advance.
95758
Let $G$ be a group, and let $H$ be a subgroup of $G$. Let $x,y\in G$ be such that $xH\subseteq Hy$. Can we conclude that $xH=Hy$? this certainly holds if $H$ is finite because both sets will have the same finite cardinality. But what about the case $|H|=\infty$? I am especially interested in the case $x=y$.  EDIT: As Thomas Andrews pointed out in the comments, it is suffices to consider the case $x=y$, because the hypothesis $xH\subseteq Hy$ implies $x\in Hy$, hence $Hy=Hx$. And so, the question becomes:     Can a subgroup (of a group, of course) strictly contain some of its conjugates?
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?