(i) Prove that the identity element of a group is unique. (ii) Prove that (a⁻¹)⁻¹ = a for every a ∈ G, a group

Question 64 of 84

Q. (i) Prove that the identity element of a group is unique.

(ii) Prove that

(a^{-1})^{-1} = a

for every

a \in G

, a group.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2018Subjective· 6mImportance★★★★★

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Concept understanding — Properties of Binary Operations (Closure, Commutative, Associative, Identity, Inverse)

Once an operation $*$ is confirmed binary on a set $S$ (so closure automatically holds), four more properties decide how "arithmetic-like" it behaves.

Commutative property. $*$ is commutative if $a*b=b*a$ for every $a,b\in S$ -- order does not matter. Ordinary $+,\times$ on numbers are commutative; ordinary $-$ is not ( $4-5\ne5-4$ ).

Associative property. $*$ is associative if $(a*b)*c=a*(b*c)$ for every $a,b,c\in S$ -- grouping does not matter, so a chain $a*b*c$ is unambiguous. Ordinary $-$ fails this too: $(4-5)-7=-8$ but $4-(5-7)=6$ .

Existence of identity. An element $e\in S$ is an identity element for $*$ if $a*e=a=e*a$ for every $a\in S$ . For $+$ on $\mathbb Z$ , $e=0$ ; for $\times$ on $\mathbb Q$ , $e=1$ .

Existence of inverse. If an identity $e$ exists, then $b\in S$ is the inverse of $a$ (written $b=a^{-1}$ ) if $a*b=e=b*a$ . For $+$ on $\mathbb Z$ , the inverse of $m$ is $-m$ ; for $\times$ on $\mathbb Q$ , the inverse of a nonzero $x$ is $\tfrac1x$ . (The notation $a^{-1}$ names an element, not the fraction $\tfrac1a$ .)

Uniqueness is guaranteed, not assumed.

Theorem 12.1 (Uniqueness of Identity). If an algebraic structure $(S,*)$ has an identity element, it has only one. Proof idea: if $e_1,e_2$ are both identities, treat $e_1$ as the identity acting on $e_2$ to get $e_1*e_2=e_2$ , then treat $e_2$ as the identity acting on $e_1$ to get $e_1*e_2=e_1$ ; comparing gives $e_1=e_2$ .

Theorem 12.2 (Uniqueness of Inverse). If $a\in S$ has an inverse, it has only one. Proof idea: if $a_1,a_2$ are both inverses of $a$ , then $a_1=a_1*e=a_1*(a*a_2)=(a_1*a)*a_2=e*a_2=a_2$ . …

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(i) Compare $e*e'$ computed two ways to force $e=e'$ . (ii) Use the defining relation of an inverse together with uniqueness of inverses in a group.

Let $(G,*)$ be a group.
(i) Uniqueness of the identity. Suppose $e$ and $e'$ are both identity elements of $G$ .
Since $e$ is an identity, $e*e'=e'$ (identity acting on $e'$ ).
Since $e'$ is an identity, $e*e'=e$ (identity acting on $e$ ).
Comparing steps 3 and 4: $e'=e*e'=e$ , so $e=e'$ . Hence $G$ has exactly one identity element.
(ii) $(a^{-1})^{-1}=a$ . Let $a\in G$ and let $a^{-1}$ denote its inverse, so by definition $a*a^{-1}=a^{-1}*a=e$ .
Read this same equation as a statement about $a^{-1}$ : it says $a^{-1}*a=a*a^{-1}=e$ , i.e. $a$ satisfies exactly the defining property required of "the inverse of $a^{-1}$ ".
In a group, every element has a unique inverse (a standard consequence of associativity: if $b$ and $c$ both invert $x$ , then $b=b*e=b*(x*c)=(b*x)*c=e*c=c$ ). …

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Q.(i) Prove that the identity element of a group is unique.

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