Suppose that we have some group G in which the following holds for all $x $ belongs to G: $x^{-1} =x$. Prove that G is abelian.  How would you guys solve this one?
123050
If $\forall x \in G$ where G is a group $x=x^{-1}$ then how is this group abelian .   This question is from Gallian's book where question provides information that G is a group and $\forall x \in G$ , $x=x^{-1}$ then we have to prove that G is abelian , how is it possible ??
16646
Suppose $a$ and $b$ are elements of a group $G$. If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.   I've been trying to prove but still inconclusive. Please prove to me.  Thanks very much for proof.
103317
Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian.  It is easy to prove it for the case of $[G' : G'']=1$ since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance.
80770
Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.
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"?
16266
If $G$ is a group show that if $(a \cdot b)^2 = a^2 \cdot b^2$ then $G$ must be abelian.   $\begin{aligned}(a \cdot b)^2 = a^2 \cdot b^2 &amp; \iff (a\cdot b)\cdot(a \cdot b) = (a \cdot a)\cdot (b \cdot b) \\&amp; \iff a \cdot (b \cdot (a \cdot b)) =a(a \cdot (b\cdot b)) \\&amp; \iff (a^{-1} \cdot a) \cdot (b \cdot (a \cdot b)) =(a^{-1}\cdot a)(a \cdot (b\cdot b))\\&amp; \iff (b \cdot (a \cdot b)) =(a \cdot (b\cdot b)) \\&amp; \iff (b \cdot a) \cdot b =(a \cdot b)\cdot b\\&amp; \iff (b \cdot a) \cdot (b \cdot b^{-1}) =(a \cdot b)\cdot (b \cdot b^{-1}) \\&amp; \iff b \cdot a = a \cdot b\end{aligned}$  Thus $G$ must be abelian. Is this right? Is there less clunky way to write it if it's?
69006
Possible Duplicate:             Let $(G,\star)$ be a group with identity element $e$ such that $a \star a = e$ for all $a \in G$. Prove that $G$ is abelian.   Ok, what i got is this: we want to prove that ab=ba, i.e. if aa=e , a=a' where a' is the inverse and bb=e, b=b' where b' is the inverse so ab=(ab)'=b'a'=ba....
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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: $a \cdot b = (b \cdot b)  \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.
82510
This is an assignment question: I just need know whether I am on the right track.....  Let $G_1,…,G_n$ be groups. Prove that the direct product $G_1×⋯×G_n$ is abelian if, and only if, each of $G_1,…,G_n$ is abelian.  I know that this question have already been discussed before, I've got two thing that I need clarification on.... Is it wise for me to start this prove by assuming that $G_1,....,G_n$ (which is for each group) are Abelian, and work toward showing that the direct product of the group is abelian secondly, is there any second part of the prove that I haven't considered..  thank you...  I'll really appreciate any 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 .
14548
I am trying to prove the above with the conditions that N is normal to G and $G/N$ is abelian.  So then for $Na,Nb\in{G/N}, NaNb=NbNa$.  But $NaNb=Nab=Nba$, so then we know $ab=ba$, so G is abelian.  So if $ab\in{G}, N=abN(ab)^{-1}=abN(ba)^{-1}=baNa^{-1}b^{-1}....$ and i'm getting lost...
67856
I'm trying to understand the following proof from Dummit &amp; Foote (pg. 137) which shows why a group of order 45 is abelian. I understand everything but the last two sentences. Why is it that $|G/C_G(P)|=1$? Also, is it always true that $G/Z(G)$ is abelian? Thank you!
17525
Let $G$ be a finite group and $f:G\to G$ an isomorphism. If $f$ has no fixed points (i.e., $f(x)=x$ implies $x=e$) and if $f\circ f$ is the identity, then $G$ is abelian. (Hint: Prove that every element in $G$ has the form $x^{-1}\cdot f(x)$.)   With the hint, I can see that for all $t\in G$, $f(t)=t^{-1}$, and since $f$ is an isomorphism, $G$ is abelian. But I can't see how to prove the hint or why it's evident.  Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50
98025
Isn't it the same to say that a group is abelian, and that the center of the group is all the group?   I have an exercise to prove that this is true, and it's exactly one stroke for each direction of the proof, correct me if I wrong:   First direction: The group $G$ is abelian, therefore for each $a,b\in G$: $$ab\:=\:ba$$ Therefore, in other words: $$a\in G\::\:Z(G)=\{ab\:=\:ba\::\:b\in G\}$$ and by definition of center (which here I symbolized as $Z(x)$), the center here is every element of $G$, therefore it is whole $G$ itself.  Second direction:  That is true: $$a\in G\::\:Z\left(G\right)=\left\{ab\:=\:ba\::\:b\in G\right\}$$ and in other words, for each for each $a,b\in G$: $$ab\:=\:ba.$$  What should I add in this proof, does it seems that i missed something? I mean it seemed pretty trivial, both directions.
12693
Is the following statement true or false:  If $G$ is a group with the property that $g=g^{-1}$ for all $g \in G$, then $G$ is abelian.   I believe it is false since I know that abelian or commutative property implies that every element in $G$ must have an inverse. Thus $g\cdot g^{-1}= g^{1}\cdot g = e$. I need someone to check my attempt.
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.
85573
if G is generated by {a,b} and ab=ba, then prove G is Abelian  All elements of G will be of the form $a^kb^j,\,\, k,j \in \mathbb{Z}^+$   so I need to get $a^kb^j = b^ja^k$