DHARMALGEBRAIC STRUCTURES 91Now we discuss the properties of a group homomorphism, that are,
(i) Preservance of identities
(ii) Protection of inverses
(iii) Shield of subgroups
Now we shall prove above facts,
(i) Let êx and êy are the identities of the groups X and Y correspondingly, then f(êx) = êy.
To prove this fact, assume x ∈ X then its image f(x) ∈ Y, so
f(x). êy = f(x) (since êy is the identity element of Y)
= f(x. êx) (since êx is the identity element of X)
= f(x). f(êx)
f(x). êy = f(x). f(êy)
⇒ êy = f(êy)
(ii) To prove the fact that for any x ∈ X, f(x–1) = [f(x)]–1, assume x ∈ X thus x–1 ∈ X
therefore
f(x x–1) = f(êx)(∴ x x–1 = êx)
= êy (∴ f(êx) = êy)
⇒ f(x) f(x–1) = êy
Conversely,
f(x–1 x) = f(êx)(∴ x–1 x = êx)
⇒ = êy (∴ f(êx) = êy)
⇒ f(x–1) f(x) = êy
That implies, f(x–1) = [f(x)]–1
(iii) Assume, (S, ) is the subgroup of (X, ) with identity element êx i.e., êx ∈ S.
Since, f(êx) = êy i.e. êy ∈ f(S). Now, for any x ∈ S, x–1 ∈ S, and f(x) ∈ f(S) and also f(x–1)
∈ f(S) ⇒ [f(x)]–1 ∈ f(S). Further for x 1 , x 2 ∈ S, x 1 x 2 ∈ S and so for f(x 1 ), f(x 2 ) ∈ f(S),
and f(x 1 x 2 ) = f(x 1 ) f(x 2 ) ∈ f(S). Hence, (f(S), ) is a subgroup of (Y, ) due to every
element of f(S) must be written as f(x 2 ) for some x 2 ∈ S.
Example 4.13 Consider two groups (R, ×) and (R+, ×) where R is the set of reals and R+ is the
set of positive reals and there is a mapping f : R → R+, i.e., f(x) = ex for all x ∈ R. Then prove
that R and R+ is not isomorphism to each other.
Sol. From the definition of group isomorphism if f is bijective (one – one and onto) and f(x × y)
= f(x) × f(y) then group R and R+ is isomorphism to each other.
l Assume x 1 and x 2 ∈ R, then f(x 1 ) = ex^1 and f(x 2 ) = ex^2. If f(x 1 ) = f(x 2 ) then ex^1 = ex^2 , it
means, x 1 = x 2. Hence, f is one – one. Also, since f(x) = ex = y (∈ R+) ⇒ log y = x or,
f(logy) = elog y = y. It means, for every element y of set R+ has a preimage log y, which
is real, hence it is in R. Thus, f is also onto.
Therefore, f is surjective.
l To prove second condition, let x, y ∈ R, so f(x × y) = exy = (ex)y, and f(x) × f(y) = ex ×
ey=ex+y. So, f(x + y) = f(x) × f(y).
Hence, we conclude that groups (R, ×) and (R+, ×) are not isomorphism to each other.