I am a bachelor student in mathematics and studying group theory at home I came across this exercise where there is need to show the existence of this group, because given any prime, is it possible to ensure that there is a group of order $p^{3}$ ?
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Can anyone helo me out with this question: "Prove that for each prime there exists a nonabelian group of order $p^3$".  This is my attempt:  I know how to show that $|Z(G)|= p$. So, since $C(x)$ contains $Z(G)$ for every $x$ not in $G$ then $|C(x)|&gt; |Z(G)|$ i.e. $|C(x)|&gt; p$. But $C(x)$ is always a subgroup of $G$ so it’s order must divide the order of $G$ and so $|C(x)= p^2$. (If $|C(x)|= p^3$ then $|G:C(x)|= |G|/|C(x)| = p^3/p^3 = 1$, so the number of elements in every conjugacy class would be 1, which would imply that every element is in the center of $G$, which in turn would mean that $G$ is abelian, which is not what we're after.) If $|C(x)|= p^2$ then $|G:C(x)|= |G|/|C(x)| = p^3/p^2= p$. Here’s where I don’t know how to go forward. I know that if there would be $p^2 – 1$ conjugacy classes then  $|G |= |Z(G)|+ (p^2-1) \cdot|G:C(x)|= p + (p^2-1)p = p + p^3 – p = p^3$. But how do I know that there actually exists a group with exactly $p^2 – 1$ conjugacy classes?  Greatful for input on my reasoning and also other ways of proving this, including how to construct such a group.
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So I've been trying to prove that there's no simple group of order $300$. This is what I did and I was wondering if it was enough.   $|G|=2^2 \cdot 3 \cdot 5^2$. Suppose $G$ is simple. Then there would be $6$ Sylow $5$-subgroups, one of which will have an index of $6$. But then $|G|=300$ does not divide $6!$ which leads to a contradiction. So, the number of Sylow $5$-subgroups is $1$ and $\exists$ a proper normal Sylow $5$-subgroup in $G$. Hence $G$ is not simple.
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I'm a bit lost on this problem:      Show that a group of order $p^2$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}$ $\times$ $\mathbb{Z}_{p}$.    The only thing I can think of that may relate to this problem is Lagrange's Theorem, where the order of a subgroup divides the order of a group. But I'm not sure if I can even use that for this problem... Any help would be tremendously helpful. Thank you!
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consider the group $G=\mathbb Q/\mathbb Z$. Let $n$ be a positive integer. Then is there a cyclic subgroup of order $n$?   not necessarily.   yes, a unique one.  yes, but not necessarily a unique one.  never
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Question :   If $|G|=pq$ where $p$ and $q$ are primes that are not necessarily distinct.   Prove that the order of $Z (G) =1$ or $pq$.   Showing the order is $pq$ is trivial.   I unsure how to start with showing the order is $1$. Hints are appreciated.  Thanks in advance.
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I know that all groups of order $\leq$ 5 are Abelian and all groups of prime order are Abelian. Are there any other examples? If so is there something special about the orders of these groups?
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If the order of elliptic group is prime then every point is a generator of that group. I tested the above statement on some elliptic curves and found it true. Does that really work on all curves?  Is there any lemma or theorem which states that?
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Let $p,q$ be primes and let $G$ be a group of order $p^2q^2$, what's the best way to show $G$ is non-simple?      I know it suffices to show that one of the Sylow-p or Sylow-q subgroup of $G$ is normal, but the counting elements argument doesn't work here since different Sylow subgroups may have non-trivial intersection.
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Show, without Lagrange theorem, that every group $G$ of order $6$ has an element of order $3$.   I know that if $G$ has an element $a$ of order 6, then $a^2$ has order $3$, but I am not shure how to show that for the case where $G$ does not have an element of order $6$. I also know that we can only have elements $1,2,3,$ or $6$ (we proved this independent without Lagrange theorem).
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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 $n$?
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Sorry for the last mistaken problem I just posted. Now I know that only having the order being odd square free is not enough for a group to be cyclic.   Here's the complete problem which the main goal is to show that any group of order $n$ satisfies $\gcd (n,\varphi(n))$=1 is cyclic. It asks me to do it in the following way:  (a) If $n &gt; 2$, $n$ is an odd squarefree integer.  (b) Show that there is a $H = G/N$ of prime order.  (c) Show that $G \cong N \times H$ and $G$ is abelian.  (d) $G$ is cyclic.    Any ideas is appreciated.
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Let $G$ be a finite group with more than one element. Show that $G$ has an element of prime order
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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.
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Jack is trying to prove:     Let $G$ be an abelian group, and $n\in\Bbb Z$. Denote $nG = \{ng \mid g\in G\}$.      (1) Show that $nG$ is a subgroup in $G$.      (2) Show that if $G$ is a finitely generated abelian group, and $p$ is prime,    then $G/pG$ is a $p$-group (a group whose order is a power of $p$).   I think $G/pG$ is a $p$-group because it is a direct sum of cyclic groups of order $p$. But I cannot give a detailed proof.
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If there exist any group which has only elements of prime order? (except of $p$-elementary abelian, $Q_{8}$ and $D_{2p}$)
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Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime.   I tried as, if $a$ and $b$ two element of $G$ of order $p$ and $q$ then $\langle a \rangle$ and $\langle b \rangle$ are two proper subgroups – a contradiction. Thus every element of $G$ have order some power of $p$. After this I am stuck. Thanks for help.
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I am to trying to prove this theorem: A finite $p$-group cannot be simple unless it has order $p$.  I have this:   Let $G = P$ and $|G|=p$; then there exists $N$, a normal subgroup of $G$ by Lagrange's theorem, such that $|N| \mid |G|$ so, $|N|= |e| = 1$ or $|N| =  p$  If $|N|=p$ so $N=G$.  But I don't know how I can prove the part in the case of $|N|=|e| = 1$  Is my start correct?