# To prove in a Group Left identity and left inverse implies right identity and right inverse

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?

• How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works. – Michael Burr Mar 21 '15 at 11:44
• @MichaelBurr please check now – Taylor Ted Mar 21 '15 at 11:45
• 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. – Michael Burr Mar 21 '15 at 11:53
• – 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 $(1)$, $$(bab)(bca)=e\implies b((ab)(bc)a)=e\implies ba=e$$

• (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? – Jason Jul 27 '17 at 8:08
• 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? – user170039 Jul 27 '17 at 13:31
• Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you! – galra Mar 3 '18 at 1:10
• @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. – user170039 Mar 3 '18 at 4:35

1.

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

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

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

2.

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