Let $G$ be a finite group and suppose that $f$ is an automorphism of $G$ with order 3 that fixes only the identity element of $G$. Show that $G$ has a single (i. e. normal) $p$-Sylow subgroup for each prime $p$ dividing $|G|$.   Been working for hours and haven't made much progress. I did note that since $f$ considered as a permutation factors as the product of disjoint 3-cycles, 3 divides $|G|$. So my next step is showing that the 3-Sylow subgroup is normal.  Edit: sorry, since $f$ fixes the identity it actually factors into a bunch of 3-cycles and a 1-cycle. So 3 does not divide $|G|$.
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.
88763
Let G be a group with $o(G)=pq$, $p&gt;q$ are primes.  Prove that $G$ has a subgroup of order $p$ and a subgroup of order $q$.  Remark: I cannot find the topic with this problem.  I have tried different methods but no one of them did not bring result. Please show how to do that. Do not use Sylow's Theorem.
120865
Show that a group with $|G| = 33$ contains an element of order $3$.   Using Lagrange's theorem, the possible orders are $1$,$3$,$11$, and $33$ (from the factors of $33$).  If there is a subgroup containing an element with an order of $33$, then we are done.   There is only one identity element with an order of $1$.   I am currently left with elements of orders $3$ or $11$. What is the best way to show that any element of order $11$ is a contradiction?
35738
Suppose that I have a group $G$ of order $p^{2}q$ for two distinct primes $p$ and $q$.  I need to first show that one of its Sylow subgroups is normal.  I start by letting $H$ be a Sylow $p$-subgroup and $K$ be a Sylow $q$-subgroup.  If $K$ is not normal, then letting $r$ denote the number of Sylow $q$-subgroups, I have that $r|p^{2}$ and $r\equiv 1$(mod $q$).  I also have by the second Sylow theorem that $r\neq 1$.  So $r = p$ or $p^{2}$.  If $r = p^{2}$, I can write out the Sylow $q$-subgroups, and count their combined elements to show that $H$ is uniquely determined and then again imply Sylow number $2$ to obtain that $H$ is normal.  But if $r = p$, then I cannot see how to proceed.  Can anyone give any advice?  Thank you.  ( I am happy to provide more detail for the $r = p^{2}$ case if it is desired or appropriate. )
20893
Question: Determine all the finite groups that have exactly one nontrivial proper subgroup.  MY attempt is that the order of group G has to be a positive nonprime integer n which has only one divisor since any divisor a of n will form a proper subgroup of order a. Since 4 is the only nonprime number that has only 1 divisor which is 2, All groups of order 4 has only 1 nontrivial proper subgroups (Z4 and D4)
98880
Let $S$ be a normal p-subgroup of a finite group $G$. Prove that $S \subseteq P$ for every Sylow p-subgroup $P$ of $G$.   Now, I know that this involves the Sylow Theorems, of course. This is very new to me but my thinking so far is this.   S is a normal p-subgroup of a finite group G, so we can say that S has some order of a power of p, $|S|=p^k$ where p is prime and S is a subgroup of some Sylow p-subgroup $K$ (by the second Sylow Th). If I let $P$ be any other Sylow p-subroup, we must show $S \subseteq P.$   Using the third Sylow Theorem, we can state that $K$ and $P$ are conjugates. Thus, there exists $x \in G$ s.t. $xKx^{-1}=P$ Thus, $xSx^{-1} \subseteq P$. But, since S is normal in G, we have $xSx^{-1}=S$.  Please let me know if this is sound theory and if not, please offer corrections.
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 $n$?
70590
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
20106
Let $p$ be a prime number . The order of a $p$-Sylow subgroup of the group $GL_{50}(F_p)$ of invertible $50\times50$ matrices with entries from the finite field $F_p$,equals which of the following    1.$p^{50}$ 2.$p^{125}$ 3.$p^{1250}$ 4.$p^{1225}$    I think order of this  group is $p^{2500}$  .A  sylow p-subgroup of order $P^k$ divides  order of the group but $p^{k+1}$  does not. I confused here how to apply these things .
109521
Let $p$ be a prime number and $G$ a finite group where $|G|=p^n$, $n \in \mathbb{Z_+}$. Show that any subgroup of index $p$ in it is normal in $G$. Conclude that any group of order $p^2$ have a normal subgroup of order $p$, but without using the Sylow theorems.
34929
The question is: If the number of Sylow $p$ groups is not congruent to $1 \pmod{p^{2}}$, then there exists two distinct Sylow $p$ subgroups $P,Q$ such that $[P : P\cap Q] = [Q : P\cap Q] = p$. I know that whenever $PQ$ is a group, $|PQ| = \frac{|P||Q|}{|P \cap Q| }$ and that there are $1+kp : 1 \leq k \leq p-1$ possible Sylow $P$ subgroups, but I do not know how to proceed.
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.
89072
Prove that a group of order 36 must have a normal subgroup of order 3 or 9.  Let n2 be the number of 2-Sylow subgroups of G (with |G|=36). Then n must be 1 or 3.  Let n3 be the number of 3-Sylow subgroups of G. then n3=1 or n3=4 if n3=1 we have 1 3-sylow group of order 9. and it is also a normal group(from sylow theorem ) if n2=1 there is normal group of order 4 but I cant show normal group of order 3.
97054
If G is a finite group and N is a normal subgroup in G , then can we say G $\cong$ G/N $\times$ N always? Is it true for like normal nilpotent or normal solvable or any such special classes.   I couldn't help but ask for it. What if G is infinite? Please tell me if you can, the concept I am skipping here. If you can avoid using exact sequences, It will be much clearer to me. I have seen an answer which still is not so clear to me. Thanks!
122933
Suppose $G$ is a group and has exactly two nontrivial proper subgroups. Prove that $G$ is cyclic and $|G|=pq$ where p,q are distinct primes or $G$ is cyclic and $|G|=p^3$ where $p$ is a prime. Usually my tactic to show a group is cyclic is find an elements such that its order and order of the group are the same. But here I cannot do so because I have no informatio regarding order of elements. So I stuck. Can anyone help me?
40359
It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).  Is an analogous statement true for all finite groups? I.e., given a finite group $A$, are there - again up to isomorphism - only finitely many groups $G$ with $Aut(G)\cong A$?  If yes, is there an upper bound on the number of such groups $G$ depending on a property of $A$ (e.g. its order)?  And if not, which groups arise as counterexamples?  And finally, what does the situation look like for infinite groups $G$ with a given finite automorphism group? And what if infinite automorphism groups $A$ are considered?
106485
Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this?