Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :

A. There exists an $e$ in $G$ such that $a \cdot e=a$ for all $a \in G$.

B. Given $a \in G$, there exists an element $y(a) \in G$ such that $a \cdot y(a) =e$.

Prove that $G$ must be a group under this product.

Attempt -Since Associativity is given and Closure also, also the right identity and right inverse is given .So i just have to prove left identity and left inverse.

Now as $ae=a$ post multiplying by a, $aea=aa$. Now pre multiply by a^{-1} I get hence $ea=a$. And doing same process for inverse Is this Right?

  • $\begingroup$ How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works. $\endgroup$ – Michael Burr Mar 21 '15 at 11:44
  • $\begingroup$ @MichaelBurr please check now $\endgroup$ – Taylor Ted Mar 21 '15 at 11:45
  • $\begingroup$ But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular. $\endgroup$ – Michael Burr Mar 21 '15 at 11:53
  • $\begingroup$ See also: math.stackexchange.com/questions/65239/… $\endgroup$ – Martin Sleziak Jan 11 '16 at 11:12

Let, $ab=e\land bc=e\tag {1}$ for some $b,c\in G$. And, $ae=a\tag{2}$ From $(2)$, $$eae=ea\implies(ab)a(bc)=ea\implies ((ab)(ab))c=ea\implies ec=ea\tag{3}$$

Similarly, $$ae=a\implies a(bc)=a\implies (ab)c=a\implies ec=a\tag{4}$$

$(3)$ and $(4)$ implies, $$ea=a$$

Also from $(3)$ and $(1)$, $$(bab)(bca)=e\implies b((ab)(bc)a)=e\implies ba=e$$

| cite | improve this answer | |
  • $\begingroup$ (1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence? $\endgroup$ – Jason Jul 27 '17 at 8:08
  • $\begingroup$ Note that given $a\in G$ there exists an element $y(a)\in G$ such that $a\cdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason? $\endgroup$ – user170039 Jul 27 '17 at 13:31
  • $\begingroup$ Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you! $\endgroup$ – galra Mar 3 '18 at 1:10
  • 3
    $\begingroup$ @galra: See the edit. Observe that by $(3)$ we have, \begin{align*}(bab)(bca)&=(be)(ea)\\&=b(ec)&\text{by (3)}\\&=(be)c\\&=bc\\&=e\\\end{align*}And by $(1)$ we have, \begin{align*}(bab)(bca)&=b(ab)(bc)a\\&=b(e)(e)a\\&=ba\end{align*} Hope it helps. $\endgroup$ – user170039 Mar 3 '18 at 4:35


$(y(a)\cdot a)\cdot (y(a)\cdot a) = y(a) \cdot (a \cdot y(a))\cdot a = y(a) \cdot e \cdot a=(y(a)\cdot e) \cdot a = y(a) \cdot a$.

$(y(a)\cdot a)\cdot ((y(a)\cdot a) \cdot y(y(a) \cdot a)) = (y(a) \cdot a) \cdot y(y(a) \cdot a)$.

$y(a)\cdot a = e$


$e\cdot a = (a \cdot y(a))\cdot a=a\cdot(y(a)\cdot a)=a\cdot e=a$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.