If (G,$\ast$) be a group and H be a subset of G which is not closed under$\ast$.So can we say by hereditary property H is closed under $\ast$. I am reading a group theory text where I found that {Set of positive Irrationals }U{1} is associative ,have a identity,every element have inverse under multiplication but due to failure of closure property it's not a group.But my question is if the set is not closed then how can associative property holds. Can we say (sqrt(2)$\cdot$sqrt(2))$\cdot$sqrt(3) and sqrt(2)$\cdot$(sqrt(2)$\cdot$sqrt(3)) are same. According to me they are not same as the first one doesn't exists as the operation is restricted on Irrationals. If we consider the group (R,$\ast$)($\ast$is multiplication operation) then $\ast$ is associative on R\Q,but the restriction mapping is not associative. So 1)Associative property does not hold if $\cdot$ is defined on R\Q×R\Q 2)Associative property holds if $\ast$ is defined on R but for the restricted map$\ast$/R-Q associative property fails. Am I true or there is any logical flaw.Please help.

  • 2
    $\begingroup$ Normally associativity is a property of a binary operation, which by definition is closed. $\endgroup$ Commented Aug 6, 2017 at 3:13
  • $\begingroup$ If you put asterisks around text, it italicizes the text instead of showing a couple asterisks. You should instead use $\LaTeX$ and type $\ast$ which will show up as $\ast$. So then you can put asterisks around text..., $\ast$like this$\ast$. Now you can fix your post. You can also use $\cdot$ for a $\cdot$ symbol $\endgroup$
    – anon
    Commented Aug 6, 2017 at 3:29

1 Answer 1


Hmmm, I'd disagree with the book.

I'd say for $a = \sqrt 2; b = \sqrt 2; c = \sqrt 3$ then $a*(b*c) = 2\sqrt 3$ but $(a*b)*c$ is undefined as $(a*b)$ is neither irrational nor $1$.

However, I suppose the book is thinking, $*$ is defined not merely in $H$ but in a larger $G$. In which case $(a*b)*c = a*(b*c)$ so it is associative.... over $G$.

I'd say it's semantic. But I agree with you, if we are talking about $*$ being associative while talking about a set $H$, I would assume we are talking about $*$ being associative over $H$. Which is absolutely is not.

.... so the book isn't wrong but... I'd say it could be argued.

  • $\begingroup$ So if (R\Q,$\cdot$) is given then whether I have to consider (R,$\cdot$) for associativity or not $\endgroup$ Commented Aug 6, 2017 at 3:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .