$(G,\cdot)$ a group. If $\exists n\in \mathbb{Z} $ such that $(a\cdot b)^{n+i}=a^{n+i}\cdot b^{n+i}$ for   $i=0,1,2.$ $\forall a,b \in G$.  Prove that $(G,\cdot)$ is Abelian. I'm not sure how to prove this. Why $i$ must be 0 ,1 or 2?
41628
I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4):     If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, show that $G$ is abelian.   I managed to prove it, but I'm not very happy with my result (I think there's a neater way to prove this). Anyway, I'm just looking to see if there's a different approach to this.   My approach:  Let $j=i+1, k=i+2$ for some $i\in \mathbb{Z}$.  Then we have that $(ab)^i = a^ib^i$, $(ab)^j = a^jb^j$ and $(ab)^k = a^kb^k$.  If $(ab)^k = a^kb^k$, then $a^jb^jab =a^jab^jb$.   We cancel on the left and right and we have $b^ja = ab^j$, that is $b^iba = ab^j$.  Multiply both sides by $a^i$ on the left and we get $a^ib^iba = a^jb^j$, so $(ab)^iba = (ab)^j$. But that is $(ab)^iba = (ab)^iab$.   Cancelling on the left yields $ab=ba$, which holds for all $a,b \in G$, and therefore, $G$ is abelian.  Thanks!
24118
If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?    Proof: $a \cdot b = (b \cdot b)  \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.
68864
How do I prove group of order 15 is abelian?  Is there any general strategy to prove that a group of particular order(composite order) is abelian?
51751
Possible Duplicate:             If $G$ is a group and $Z(G)$ the center of $G$, show that if $G/Z(G)$ is cyclic, then $G$ is abelian.  This is what I have so far:  We know that all cyclic groups are abelian. This means $G/Z(G)$ is abelian.  $Z(G)= \{z \in G \mid zx=xz \text{ for all } x \in G \}$.  So $Z(G)$ is abelian.    Is it sufficient to say that since $G/Z(G)$ and $Z(G)$ are both abelian, $G$ must be abelian?
109512
So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $G$ is abelian iff $[G,G] = {e}$. Where $e$ is the neutral element of $G$ and $[G,G]$ is the commutator group of $G$.  Let $H\triangleleft G$, Show that $G/H$ is abelian iif $[G,G] \subseteq H$.  I've proved this direction  $\Rightarrow $) If $G/H$ is abelian, $\forall x,y \in G$,  $xHyH = xyH = yxH = yHxH$.   Therefore have that $G$ is abelian. As shown earlier, $[G,G] = {e}$ if $G$ is abelian. So $[G,G] = {e} \subseteq H$.  I can't seem to figure out the other direction...  $\Leftarrow$)  I only have that for $x,y \in G$,  $x^{-1}y^{-1}xy \in H$ and that $H\triangleleft G$. It does not seem to lead me anywhere.  Thanks for the help!
100258
Let $G$ be an Abelian group of order $mn$ where $\gcd(m,n)=1$.   Assume that $G$ contains an element of $a$ of order $m$ and an element $b$ of order $n$.   Prove $G$ is cyclic with generator $ab$.    The idea is that $(ab)^k$ for $k \in  [0, \dots , mn-1]$ will make distinct elements but do not know how to argue it.   Could I say something like $&lt;a&gt;=A$, $&lt;b&gt;=B$, somehow $AB=\{ ab : a \in A , b \in B \}$ and that has order $|A||B|=mn$?  Don't know if it's the same exact or similar to .
109914
Let $G$ be a group. If for all $a,b,c\in G$, $ab=ca\Rightarrow b=c$, then phow can I prove that  $G$ is abelian?
112845
Let $ G $ be an abelian group of order 6.  Then $ G $ has one element of order 6.  And so, $ G $ is cyclic and isomorphic to $ \mathbb Z _6 $.  In general, It is not true that an abelian group of order $n$ has an element of order $n$.  However in the case of $n=6$, it is true. I'm wondering there is any rule of the case (for example, a rule of $n$ that makes an abelian group cyclic).
4434
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.   This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.   (Please note that $e$ in the question is the group's identity.)  Here's my attempt though...  First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...  So, I begin by trying to play around with the elements of the group based on their definition...  $$(g_2g_1)^r=e$$  $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$  I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,  $$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$  Then ultimately...  $$g_1g_2=e$$  I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.  Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra.
69532
What is the intuition behind the fact that if every element in a group is of order $2$, we have that the group is abelian? I can prove it, but I do not know the intuition behind it.
107457
Since $105 = 3\cdot5\cdot7$,  implies that there must exist a normal $7$-Sylow subgroup. Does this help prove that $G$ is abelian? Also, can this result be generalized (If $|G| = p\cdot q\cdot r$, and $G$ has a normal $p$-Sylow subgroup, then $G$ is abelian (where $p&lt;q&lt;r$).
27569
I am looking for an example of a group $G$ where the equality $(ab)^{n}=a^{n}b^{n}$ holds for two consecutive integers $n$, but $G$ is not an abelian group. I've started do some calculations in the group $D_{4}$ (I gave up!) Do you know where I can find that example? Is it possible find such example without doing a lot of calculations?  Thanks for your help!
74376
I'm currently reviewing my algebra for my last prelim and came across the following problem that has me stumped:  If $A,B,C $ are finite Abelian groups such that $A\oplus B \cong A\oplus C$ then show that $B\cong C$.  I'm not really sure where to start; the only thing that comes to mind is the fundamental theorem of finitely generated Abelian groups, I'm not sure if that's useful though.   I really only want a good hint.
27740
I'm stuck at this question. Can someone please help me?     Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.   Let $x$ be the element of $G$ which has order 2. Let $y$ be an arbitrary element of $G$. We have to prove that $x \cdot y = y \cdot x$.  Since $x$ has order $2$, \begin{equation} x^2 = e \end{equation} That is, \begin{equation} x^{-1}=x \end{equation}  I don't really know how to proceed. I've tried a number of things, but none of them seem to work.
12037
Continuing my work through Dummit &amp; Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in ):     Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian. [If $G/Z(G)$ is cyclic with generator $xZ(G)$, show that every element of $G$ can be written in the form $x^az$ for some $a \in \mathbb{Z}$ and some element $z \in Z(G)$]      The hint is actually the hardest part for me, as the quotient groups are somewhat abstract. But once I have the hint, I can write:  $g, h \in G$ implies that $g = x^{a_1}z_1$ and $h = x^{a_2}z_2$, so \begin{align*}gh &amp;= (x^{a_1}z_1)(x^{a_2}z_2)\\\  &amp;= x^{a_1}x^{a_2}z_1z_2\\\ &amp; = x^{a_1 + a_2}z_2z_1\\\ &amp;= \ldots = (x^{a_2}z_2)(x^{a_1}z_1) = hg. \end{align*} Therefore, $G$ is abelian.  1) Is this right so far?  2) How can I prove the "hint"?
104025
I am a student trying to learn some abstract algebra this summer, and I recently proved (as an exercise) that if $G$ is a group where every element has order 2, then $G$ is abelian. I was wondering could we make a similar conclusion about groups where every element has order 3, namely I am asking if $G$ is a group where all elements have order 3, then $G$ is abelian. I think that this is not true, but I cannot think of a counterexample.   The only groups that I can think of which have all elements order 3 are the groups $(\mathbb{Z}/3\mathbb{Z})^n$, but these are abelian. Any help is appreciated. Thanks!
84244
If $G$ is a group in which $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.  Proof: Let $x$ be the smallest of the 3 consecutive integers. Then, we have   $(1)(a\cdot b)^{x} =a^x\cdot b^x$, $(2)(a\cdot b)^{x+1} =a^{x+1}\cdot b^{x+1}$ and $(3)(a\cdot b)^{x+2} =a^{x+2}\cdot b^{x+2}$.  Using $(2)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $baba...ba = a^{x}b^{x}(4)$ whereby there are $x$ number of $a$ and $x$ number of $b$ on both sides.  Using $(1)$ and multiplying both sides on the right by $ab$, we have $\overbrace{abab...ab}^{(x+1)ab} = (a^{x}b^{x})ab (5)$.  Substitute $(4)$ into $(5)$, we get $\overbrace{ab...ab}^{(x+1)ab} =\overbrace{ba...ba}^{(x)ba}ab(6)$.  Using $(3)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $\overbrace{ba...ba}^{(x+1)ba} = a^{x+1}b^{x+1} = \overbrace{abab...ab}^{(x+1)ab}(7)$.  Combining $(6)$ and $(7)$, and multiply $a^{-1}b^{-1}...a^{-1}b^{-1}$ on the left, we get $ab = ba$. Hence $G$ is abelian.