Question

problem A-18 Find all points (ary) where the
functions fo), gea), h(x) have the same value.
fox) = 2 +3 g(x) = 2x-5, h(c) 2 2 + 10​

Answer 1

Answer:

Suppose for contradiction that gx1 = gx2. Since G is a group, g

−1 ∈ G. So this

means that g

−1

(gx1) = g

−1

(gx2). By associativity, this means that (g

−1

g)x1 =

(g

−1

g)x2. This simplifies to ex1 = ex2, where e is the identity. Finally, by the

property of the identity, we get that x1 = x2. But this contradicts the assumption

that x1 6= x2. So we have shown that if x1 6= x2 then gx1 6= gx1. Thus all the

elements of the form gx are distinct.

Similarly, we have to show that if x1 6= x2 ∈ G then x1g 6= x2g. Again, suppose

not. That is, suppose that x1g = x2g. But then when we multiply both sides by

g

−1 on the right, and use the same group properties as above, we get that x1 = x1.

Again, this is a contradiction, so we must have that all elements of the form xg are

distinct.

Next we have to show that the sets S = {gx|x ∈ G} and S

0 = {xg|x ∈ G} fill

out G. That is, for each element h ∈ G, we need to find elements x, x0 ∈ G s.t.

xg = gx0 = h. So let x = hg−1 and let x

0 = g

−1h. We know that x, x0 are in G

since g

−1 ∈ G by the inverse property, and the products are in G as G is closed

under multiplication.

Now we just compute:

xg = (hg−1

)g

= h(g

−1

g)

= he

= h,

and similarly we can compute that gx0 = g(g

−1h) is just h after using all three of

the group properties.

So for each element h ∈ G, we have found x, x0

s.t. xg = gx0 = h. Therefore the

sets S and S

0 fill out G.