Prove if the following are abelian groups, if so, list the neutral element e:

(a) $G:= \Bbb Z, a*b:= max \{a,b\}$ (Z as in positive integers incl. zero (i dont know how to write it with mathjax))
(b) $G:= \Bbb R, a*b:= a^b$ (only positive real numbers)
(c) $G:= \Bbb Z, a*b:= lcm(a,b)$ (least common multiple of a and b) (integers greater than zero)
(d) G is the open interval $(0,1) \subset \Bbb R $ with $$x*y: = \frac{xy}{1-(x+y)+2y} $$

Hi guys, the university only gave me some definitions, no further explaination whatsoever, im not even sure i understand this correctly (uni is really terrible at teaching)

First, take example a), I assume "$G:= \Bbb Z, a*b:= max \{a,b\}$ " means a group G with the elements a and b $\subset \Bbb Z $. So i need to use the 4 definitions to prove this to be a abelian group.

EDIT: 2) (b) $G:= \Bbb R, a*b:= a^b$

Associativity: $\forall a,b,c \in G: a\circ(b\circ c) = (a\circ b)\circ c$ For question b would mean $a^{b^c}=a^{b^c}$ which is correct...

Invertibility:$\forall a \in G \exists a' \in G: a' \circ a = e$ Suppose its invertible -> $a'\circ a=e \rightarrow a^{-1^{a}}=e$ However, i do see from my schools paper that $a \circ a^1=e$ should also hold, but then in this case that would be $a^{a^{-1}}$ which is not the same thing, so invertibility also fails?

Identity: $\forall a \in G: e \circ a = a$ Since invetibility fails i cant say much about identity...

Commutativity: $a \circ b = b \circ a$ $\rightarrow a^b=b^a$, unless a=b=1, that is not true either...

If someone has time to look it over and give me some suggestions or point out where is wrong or how to write the proofs more formally id be really thankfull!!!!

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    $\begingroup$ If you think that your university's teaching is hard, one book I would recommend is John E. Gallian Contemporary Abstract Algebra. It is practically written for dummies (like me). $\endgroup$ Commented Oct 4, 2021 at 8:23
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    $\begingroup$ Are you really sure that an actual mathematics professor used $\mathbb{Z}$ to denote the positive integers? I know that notations can vary, but I have never, ever, seen anyone use $\mathbb{Z}$ for anything other than the relative integers. $\endgroup$ Commented Oct 4, 2021 at 8:30
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    $\begingroup$ @CaptainLama he didnt teach that,,i just dont know how to use latex to write positive integers,,, $\endgroup$ Commented Oct 4, 2021 at 8:48
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    $\begingroup$ @AyamGorengPedes unfortunetely i wouldnt have the time, they cover this extremely quickly, 2 lectures and move on $\endgroup$ Commented Oct 4, 2021 at 8:49
  • $\begingroup$ @JerryCohen, the good thing about the book is it's simplicity. An hour with the first chapter about Groups will do wonders. A lot of Abstract Algebra is understanding the basic Axioms first, then merely expanding from there. $\endgroup$ Commented Oct 4, 2021 at 8:54

1 Answer 1


I think you are confusing the operation (which is often called multiplication for Groups). When we say we multiply 2 elements $a$,$b$ in a group, we mean we do the group operation "$\circ$" for $a$ and $b$. That is, $a \circ b$.

For example, The real numbers form an abelian group $G$, with operation $a \circ b = (a+2)+(b+2)$ (regular addition as a part of the operation)

  1. Associativity: $(a \circ b) \circ c = \left((a+2) + (b+2)\right) \circ c = (a+b+4) \circ c = (a+b+4)+(c+2) = a+b+c+6 = a \circ (b+c+4) = a \circ \left((b+2)+(c+2) \right) = a \circ (b \circ c)$.
  2. Identity: Does there exist an element $e$, such that $a \circ e = a$? Let $a$ be chosen randomly from the elements of $G$, and let $x=-4$. Notice that $a \circ x = (a+2) + (x+2) = (a+2) + (-4+2) = a$. Thus we can deduce that $e=-4$.
  3. Invertibility Does there exist an element $a^{-1}$, such that $a \circ a^{-1} = e$? This is the same as solving $(a+2)+(a^{-1}+2)=-4$, which gives us $a^{-1}=-a-8$. Once more, since we can choose $a$ randomly (or "arbitrarily") from the elements of $G$, that means all elements of $G$ has an inverse!

It can also be shown that our group $\left(G,\circ\right)$ is commutative, however I don't want to put it there, because it is not necessary to form a group. The one thing I want to emphasize is that we are not multiplying (unless the operation is a literal multiplication) anything, but rather using the group operation. In your question, you apply the group operation first, then multiply it again.

If you would like, I can also point out why/why not (a),(b),(c),(d) are groups.

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    $\begingroup$ aha so its similiar to relations $\endgroup$ Commented Oct 4, 2021 at 8:55
  • $\begingroup$ So question a wouldnt hold as a group? since $a\circ (b \circ c) $is not equal to $(a\circ b)\circ c$. Because a might be greater than b but there is no way to determine a is greater than c? $\endgroup$ Commented Oct 4, 2021 at 9:02
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    $\begingroup$ @JerryCohen for question (a), no matter the order of operation, we can guarantee that by doing $a \circ b \circ c$, the answer is the max(a,b,c). However it is not a group due to another reason! Suppose it has an identity $e$. Let us check it's invertibility. Assume it is invertible, max($a$,$a^{-1}$) = $e$. Then, either $e=a$ or $e=a^{-1}$. If the former, then every element of the group is an identity (wrong! only one identity in a group). If the latter, then $e > a$. But then, max($a$,$e$) = $e$, not $a$! (remember that by definition identity is the element such that $a \circ e=a$ $\endgroup$ Commented Oct 4, 2021 at 9:14
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    $\begingroup$ Thanks a became very clear ! If you have time could you have look at my edits? Thanks $\endgroup$ Commented Oct 5, 2021 at 7:11
  • $\begingroup$ It is not commutative yes, that is enough to show that it is not an Abelian group. And yes, for any element $a \in G$, $e \circ a = a = a \circ e$ $\endgroup$ Commented Oct 5, 2021 at 7:31

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