(Post has been edited )  So I am reading the proof of this statement:  Let G be a group of order $pq$ with $p,q$ primes and $p&lt;q$.   If $p \nmid (q-1)$, then $G\simeq \mathbb{Z}_{pq}$  if $p \nmid (q-1)$ then thre are two isomorphism classes: $\mathbb{Z}_{pq}$ and a certain non-abelian group.  So, here is part of the proof (until the part that I have problems with):  Supposing that $p \mid (q-1)$, let $P \in Syl_p(G), Q \in Syl_q(G)$, since $\mid G:Q\mid = p$, which is th least prime number that divides $\mid G \mid$, then $Q \unlhd G$. Because $(\mid P \mid, \mid Q \mid)=1$, then $P \cap Q = \{e\}$, so $G=QP$.   Then $G$ is a semidirect product $Q \rtimes_{\theta} P$ for a certain morhpism $\theta : P \to Aut(Q)$.   Now they assert that $Aut(Q) \simeq \mathbb{Z}_{q-1}$.  Why is this? I am pretty sure this is not true for any group since complete groups are isomorphic to their automorphim group by definition.  Any help would be appreciated.
841
I don't know about the Sylow Theorems.  But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.  If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$.   But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.  I am asking for help completing this argument.  Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).
35133
There is a proposition in Beachy and Blair's Abstract Algebra that I don't entirely follow. The proposition is the following:   Let $G$ be a group of order $pq$, where $p &gt; q$ are primes.  a) If $q$ is not a divisor of $p-1$, the $G$ is cyclic.  b) If $q$ is a divisor of $p-1$, then either $G$ is cyclic or else $G$ is generated by two elements $a$ and $b$ satisfying the following equations: $$a^p = e, \\ b^q = e,\\ ba = a^nb $$ where $n \not \equiv 1  \ (mod \ p)$ but $n^q \equiv 1 \ (mod \ p)$.  Can one of you prove how this is true? I understand a similar proof for when $q=2$, but this one is more complicated.
69131
I'm considering finite groups only. Cayley's theorem says the a group $G$ is isomorphic to a subgroup of $S_{|G|}$. I think it's interesting to ask for smaller values of $n$ for which $G$ is a subgroup of $S_n$. Obviously, it's not always possible to do better than Cayley's theorem. But sometimes it is possible (for example, $\mathbb{Z}_6$ as a subgroup of $S_5$).  So I'm asking:        Given a finite group $G$, is there an algorithmic way to find or approximate the minimal $n$ for which $G$ is isomorphic to a subgroup of $S_n$?   If the answer to $(1)$ is not known, is it known for specific classes of groups?   In particular, for finite abelian groups, is it true that for a prime $p$, the minimal $n$ for $\mathbb{Z}_{p^{t_1}} \times \mathbb{Z}_{p^{t_2}}$ is $p^{t_1}+p^{t_2}$ (I can prove that is is true for different primes $p_1$ and $p_2$, but have problems when it's the same prime in both factors).      Thanks!
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?
76903
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.
71935
When people want to classify a group with certain (small) order, they seem to find a normal subgroup $H$ and a subgroup $K$, and then consider how many distinct $H \rtimes_\varphi K$ are there.  My question is: For any group $G$, does it always exist normal subgroup $H$, subgroup $K$ and $\phi : K \rightarrow Aut(H)$ s.t. $G\cong H \rtimes_\varphi K $?  Sorry if this question is too basic...  Thanks the comment and the answers. Then if the $G$ is not simple, can we have can $G\cong H\rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal?  EDIT:  If we have a group $G$, that is not simple (then $G$ has at least one normal subgroup), are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible).  But the quaternion group and $Z_4$ given in answers have shown that we cannot always have it.
120637
Here's a question I've been asked;  Given distinct primes $p,q,r$, show that any group $G$ of order $pqr$ is not simple.   So far, my idea has been to individually check each possible proper subgroup, since by Lagrange there are only $$\dbinom{3}{2} + \dbinom{3}{1} + 1$$ of them, and show that at least one of them other than the trivial group is normal. Surely there is a smarter way, since generalizing this approach wouldn't work so well if I wanted to show similar results for groups who's order is a product of $n$ distinct primes.  Is there some other approach? Thanks for any insights
90920
If $G$ is an abelien group, let $\hat{G}$ be the set of all   homomorphisms of $G$ into the group of nonzero complex numbers under   multiplication. If $\phi_1, \phi_1\in \hat{G}$ define $\phi_1\cdot  \phi_2$ by $(\phi_1\cdot \phi_2)(g)=\phi_1(g)\phi_2(g)$ for all $g\in  G$.      If $G$ is a finite abelian group prove that $o(G)=o(\hat{G})$ and $G$   is isomorphic to $\hat{G}$.   The case when $G$ is finite cyclic is easy because in this case $\hat{G} \cong G$. Let's consider the case when $G$ is abelian group of order $p^n$, $p$ - prime.   By Fundamental theorem on finite abelian groups we have the following: $$G=\langle a_1 \rangle_{p^{n_1}}\times \langle a_2 \rangle_{p^{n_2}} \times \dots \times \langle a_k \rangle_{p^{n_k}},$$  where $n_1\geq n_2\geq \dots \geq n_k&gt;0$ and $n_1+\dots+n_k=n$.   Let $A_i=\langle a_i \rangle_{p^{n_i}}$ and $A_i$ are cyclic groups then $A_i\cong \hat{A_i}$. So we have the following $$G=A_1\times \dots \times A_k\cong \hat{A_1}\times \dots \times \hat{A_k}.$$  Then I was trying to think about $\hat{G}$ and I think that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$ which follows that $G\cong \hat{G}$.   How to prove rigorously that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$? I am not able to prove it.
20389
I'm trying to prove that Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$ (p prime).  I know that Aut $\mathbb Z_p$ has $p-1$ elements because $\mathbb Z_p$ has $p-1$ possiblities of generators, so intuitively I see Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$, but I couldn't prove it formally.  I'm trying to build an isomorphic function, but I don't how to do it.  Am I in the right way? Maybe I'm forgetting some trick or something.  Thanks
97103
$(G,*)$ is an abelian group and $a,b$ are elements of G. Let $n=ord(a)$ and $m=ord(b)$ in $G$. Show that the order of $a*b$ is equal to $nm$ if $gcd(n,m)=1$.  I have already proved that $ord(a*b)|nm$. But the other part is a problem.  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?
72696
This is problem 3 from Hungerford's section about the Sylow theorems.  I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. This is what I have so far:  Suppose $|G| = p^n$. For $k = 0$, $\{e\}$ is a normal subgroup of order $p^0$. Suppose $N_1, ..., N_k$ are normal subgroups of $G$ with orders as described (ie. $|N_i| = p^i$) and $k &lt; n$.  $G/N_k$ is a $p$-group, so it has a non-trivial centre $C(G/N_k)$. $\pi^{-1}(C(G/N_k))$ is a subgroup of $G$, where $\pi : G \to G/N_k$ is the quotient map. It is normal in $G$ because $\pi$ is a homomorphism and $C(G/N_k)$ is a normal subgroup of $G/N_k$.  I know it contains $N_k$ and some stuff not in $N_k$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal.
123087
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.
65018
Suppose $G, H$ are finite abelian groups with the same number of elements of any given order. Show they are isomorphic.   Since finite abelian groups are isomorphic if and only if their Sylow subgroups are, we may restrict our attention to the case where $G, H$, are $p$-groups, but I can't quite make it past there...
2057
Respected all  Please help me on this problem.   Suppose that $G$ is given group and $N$ be a normal subgroup of $G$ and $N$ is isomorphic to $G_1$ as groups. If $G/N$ is isomorphic to the group $G_2$, can we establish that $G$ is isomorphic to the group $G_1\times G_2$ ?
29009
I want to prove the following:          Let $G$ be a group of order $2^nm$, where $m$ is odd, having a cyclic Sylow $2$-subgroup.     Then $G$ has a normal subgroup of order $m$.      ATTEMPT:  We will show that $G$ has a subgroup of order $m$. Let $\theta:G\to \text{Sym}(G)$ be the homomorphism which is defined as $$  \theta(g)=t_g,  \quad  \forall g\in G $$ where $t_g:G\to G$ is defined as: $$  t_g(x)=gx,  \quad  \forall x\in G. $$ Let $g$ be an element of order $2^n$ in $G$. (There exists such an element since $G$ has a cyclic Sylow $2$-subgroup.)  The cyclce representation of $t_g$ is a product of $m$ disjoint cycles each of length $2^n$. Therefore $t_g$ is an odd permutation.  Thus the homomorphism $\epsilon\circ \theta:G\to\{\pm 1\}$, where $\epsilon:\text{Sym}\to\{\pm 1\}$ is the sign homomorphism, is a surjection.  By the First Isomprphism Theorem, we conclude that the kernel $K$ of $\epsilon\circ \theta$ is of order $2^{n-1}m$.  If $n$ were equal to $1$ then we are done.  If $n&gt;1$ then note that any Sylow $2$-subgroup of $K$ is also cyclic. This is because each Sylow $2$-subgroup of $K$ is contained in a Sylow $2$-subgroup of $G$, where the latter is cyclic.  Now we can inductively show that $K$ has a subgroup of order $m$.    What I am struggling with is showing the normality.  Can anybody please help me with this.  Thanks.
66503
Let $\mathbb Q$ be group of rational numbers under addition.  Then I want to find out the automorphism group of $\mathbb Q$. In a book it is given that Aut($\mathbb Q$) is isomorphic to $\mathbb Q^*$, the  multiplicative group of rational numbers. I stuck in this.