How do I show that $|G|=33$  has an element of order $3$. I could use Cauchy's and say that since $3$ is prime divisor of $33$ there must be a subgroup of order 3. But this problem shows up in Gilligan's book before Cauchy's theorem is introduced. How do I prove this without Cauchy's.   Edit: we also haven't been introduced to Sylow, normal or quotient groups.
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?
35999
I wanted to share with you my resolution of this exercise.     How many different groups of order $15$ there are?   My resolution:     We're looking for groups such that $|G|=15=3\cdot 5$. Then: $G$ has an unique Sylow $3$-subgroup of order $3$ ($P_3$) and an unique Sylow $5$-subgroup of order $5$ ($P_5$).      Furthermore, $$|P_3 \cap P_5|\  \Large\mid\ \normalsize|P_3|,\ |P_5|$$   So $$|P_3 \cap P_5|=1\implies P_3 \cap P_5=\{e\}.$$      Then:$$|P_3\cdot P_5|=\displaystyle\frac{|P_3|\cdot |P_5|}{|P_3 \cap P_5|}=|P_3|\cdot |P_5|=3\cdot 5=|G|,$$      and $$G=P_3\cdot P_5 \simeq P_3 \times P_5 \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}.$$   It's everything correct and well justified? There's some things that I didn't say, for example that a subgroup a prime $p$ as order is cyclic, and thus isomoprhic to $\mathbb{Z}/p\mathbb{Z}$. Maybe I did something wrong. I would appreciate someone to look and criticize my exercise.  Thank you.
50526
How many normal subgroups does a non-abelian group $G$ of order $21$ have other than the identity subgroup $\{e\}$ and $G$?  1) 0  2) 1  3) 3  4) 7  I think option 1 is incorrect because every group of order 1 to 59 is not simple. Hence group of order 21 has atleast one normal subgroup.   Now from the Sylow's theorem we get there has one subgroup of order 7 and one subgroup of order 3 or 7 subgroup of order 3. Since 3 and 7 are primes then any group of order 3 and 7 is cyclic. hence normal. I think I am wrong some where but I am unable to locket the point. please help me to choose the right 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.
115013
Let $G$ be a group of order $150$.  I must show that it has a normal subgroup of order $25$.  The hint says to show that is has a normal subgroup of order $5$ or $25$.  Now from Sylow, I know that the number $n_5$ of Sylow-$5$ subgroups (which each have $25$ elements) must be either $1$ or $6$, since $n_5$ must also divide $6$.  Now clearly if $n_5=1$ we are done, so I can assume that $n_5=6$.  My problem is I haven't figured out how to use this information (I am going for a contradiction, just based on what the problem asks me to prove).  I know by Cauchy's theorem I can get a subgroup of order $5$, but I don't see immediately if it must be normal.   Any direction I should try to be moving?
88472
Let $G$ be a finite abelian group, then $G$ has a subgroup of order $n$ if and only if $n\mid G$.  Proof: by Lagrange if $H\leq G$ then $|H|$ divides $|G|$ so this proves one of the implications.  We prove the other implication by strong induction on the order of $G$, for primes this is just Cauchy, so we can take these like base cases. Now let $n$ be a proper divisor of $|G|$, take $p$ a prime in the factorization of $|G|$ (instead of $|G|$ is should be $n$, I had made a mistake when typing the first time), then by cauchy there is an element $x$ of $G$ of order $p$, since $G$ is abelian $\langle x\rangle$ is normal in $G$ so we can look at the quotient group $G/\langle x\rangle$, this group has order $\frac{|G|}{p}$, now take a prime $p_1$ in the factorization of $\frac{|G|}{n}$. By induction the quotient group has a subgroup $E$ of order $G/pp_1$ now look at the preimage of $E$ under the natural homomorphism from $G$ to $G/\langle x\rangle$, it is easy to see this is a subgroup of $G$, we call it $I$, consider the restriction of the natural projection to $I$, this is a surjective homomorphism from $I$ to $E$ with kernel $\langle x\rangle$ so by the first isomorphism theorem $I/x\cong E$ so $\frac{|I|}{|\langle  x\rangle|}=|E|\implies  |I|=\frac{|G|}{pp_1}{p}=\frac{|G|}{p_1}$ Notice $\frac{|G|}{p_1}$ is a multiple of $n$ since $p_1$ was in the fatorization of $\frac{|G|}{n}$. So by induction $I$ has a subgroup of order $n$ which is also a subgroup of $G$.
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 $n$?
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.
124249
I need to know if every group whose order is a power of a prime $p$ contains an element of order $p$? Should I proceed by picking an element $g$ of the group and proving that there is an element in $\langle g \rangle$ that has order $p$?
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.
87453
These questions are really bothering me. Any help would be much appreciated.  Let $p$ be prime  a) If $(G, \oplus)$ has order $2p$, prove that every proper subgroup of $(G, \oplus)$ is cyclic  b) If $(G, \oplus)$ has order $p^2$, prove that $(G, \oplus)$ has a subgroup of order $p$
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.
65789
I need help to prove the following result   Let $p$ be a prime. If a group has more than $p − 1$ elements         of order $p$, why can’t the group be cyclic?   Thanks for the help.
3293
I am working on the following problem from group theory:     If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.   That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$.
116544
I have done part $a$ and $b$, stuck on $c$. I don't know what to do here. I know that since $G$ has order $p^2$, any  element must have order $1, p$ or $p^2$ so any generator must have one of those orders, but I don't know how to reason any further..
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?
78456
I have a lot of difficulty understanding this proof we went over in class about classification of groups of order 12.  Let $G$ be a finite group of order $n=p^rm$ where $m \nshortmid p$.  Denote $Syl_p(G)$ as the set of all sylow p-subgroups, and $n_p(G)$ as the cardinality of $Syl_p(G)$.  (1) G has subgroups of order $p^r$. (How do we know this??)  (2) We know all sylow p-groups are conjugate and their number $n_p(G) | m$, by the 2nd and 3rd Sylow Theorems.  (3) We have that $n_p(G) \equiv 1(p). $  Then $n=12=2^2*3^1$ so $n_2(G)|3$ gives $n_2(G) \in \{1,3\}$ and $n_3(G)|4$ so $n_3(G) \in \{1,4\}$ since $2 \not\equiv 1(3)$.  Case 1: $n_3(G) = 4.$ Then the action of G by conjugation on $Syl_3(G)$ gives a homomorphism $f: G \rightarrow S_{Syl_3(G)} \simeq S_4$. $Ker(f)$ consists of $g \in G$ that normalizes all 3-Sylow subgroups. (Why?). Let $P_3$ be a 3-Sylow subgroup. Then the order of $P_3$ is 3. (I don't get that either), and $[G:N_G(P_3)] = 4.$ Thus $P_3 \subset N_G(P_3)$ gives $P_3 = N_G(P_3)$. (I'm lost here...) So $Ker(f) = \cap Syl_3(G) = \{e\}$ (the intersection of the sylow 3-subgroups is trivial. (How did we arrive here...) Thus we conclude that $G$ is isomorphic to a subgroup of $S_4$ of order 12.  To show $G$ is isomorphic to $A_4$, we have that $G$ has 8 elements of order 3: 4 3-sylow subgroups each has 3-1=2 elements of order 3. And $S_4$ has 8 3-cycles meaning f(G) contains all 3-cycles and hence the group they generate, which is $A_4$, so $A_4 \subset f(G)$. But since $|A_4| = 12 = |f(G)$ then $f(G) = A_4$ and $G \simeq A_4$. (Not understanding the first half of the proof, I do not get this part either).  Case 2: $n_3(G) = 1.$ The scope of the question is too large, I will need to post separately to understand the second half, unless the organizational rules of math.stackexchange would require that I post here. In that case, I'll edit the question.  I basically cannot follow the proof because it seems to skip too many steps; it would be helpful if anyone could elaborate case 1 with more reasoning so that I can follow it. Any help would be appreciated; I've spent hours trying to look up and decipher the proof but without any success.