Question

यदि a , b समूह G के दो विभिन्न अक्पद है तो H. G का उपसमूह है तब H = H , पदि और केवल पदि ab CH ​

Answer 1

SOLUTION

TO PROVE

If a, b are two distinct arbitrary elements of a group G and H is a subgroup of G if and only if

$\sf{a \: , \: b \in \: H \: \implies \: a{ b }^{ - 1} \in \: H}$

PROOF

Let H be a subgroup of G

Since H is a subgroup of G

$\sf{b \in \: H \: \implies \: { b }^{ - 1} \in \: H}$

$\sf{ \therefore \: a \in \: H\: , \: b \in \: H \: \implies \: a{ b }^{ - 1} \in \: H}$

Conversely let H be a non empty subset of G such that

$\sf{a \: , \: b \in \: H \: \implies \: a{ b }^{ - 1} \in \: H}$

Let a ∈ H

$\sf{a \: , \: a\in \: H \: \implies \: a{ a }^{ - 1} = e \in \: H}$

Thus H contains identity element

Now

$\sf{e \: , \: a\in \: H \: \implies \: e{ a }^{ - 1} = { a }^{ - 1} \in \: H}$

Thus the inverse of each element in H exists in H

Let a ∈ H , b ∈ H

$\sf{ \therefore \: a \in \: H\: , \: {b}^{ - 1} \in \: H \: }$

$\sf{ \implies \: a {({ b }^{ - 1} )}^{ - 1} \in \: H}$

$\sf{ \implies \: a b \in \: H}$

$\sf{ \therefore \: a \in \: H\: , \: b \in \: H \: \implies \: a{ b }^{ } \in \: H}$

Thus H is closed

Since H is a non empty subset of G and associative property holds in G

So associative property holds in H

Therefore H is a group

Hence H is a subgroup of G

Hence the proof follows

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