Question

show that (gh)^-1 = (h^-1)(g^-1)

Answer 1

True :)

Answer 2

1 = 1

Answer 3

Or, 1/gh= 1/hg

Answer 4

yeah. but how do i show it?

Answer 5

(gh)^-1 = 1/(gh)
= (1/g) * (1/h)
= (g^-1)(h^-1)
= (h^-1)(g^-1)
= RHS

Answer 6

oh great
thanks!

Answer 7

oh heck no

Answer 8

>.>

Answer 9

this
\[f^{-1}\] means the inverse function, not the reciprocal

Answer 10

True, True, :)

Answer 11

you are trying to show that
\[(h\circ g)^{-1}=g^{-1}\circ h^{-1}\]

Answer 12

it is always true that
\[(ab)^{-1}=b^{-1}a{-1}\] now matter what a and be are. functions, anything. check as follows

Answer 13

first of all what does
\[(h\circ g)^{-1}\] mean? it means if you compose this with
\[(h\circ g)\] you should get the identity function back

Answer 14

in other words you know that
\[(h\circ g)^{-1}\circ (h\circ g) = I\]

Answer 15

oh but h and g are elements of a group. it's not a function.

Answer 16

so now lets try to show that
\[(h\circ g)^{-1}=g^{-1}\circ h^{-1}\] by checking that it is the right thing
\[g^{-1}\circ h^{-1}\circ h \circ g =g^{-1}eg=g^{-1}g=e\]\]

Answer 17

fine a group it is and i got rid of the circle notation and used e as the identity. same idea exactly

Answer 18

point is if the group is not commutative, you cannot switch around the inverses. so
\[(ab)^{-1}=b^{-1}a^{-1}\]

Answer 19

easy in this notation anyway.
\[b^{-1}a^{-1}ab=b^{-1}eb=b^{-1}b=e\]

Answer 20

and likewise on the other side

Answer 21

oh ok

Answer 22

by the way they could be functions right? permutations, anything. that is why i said above that this is completely general

Answer 23

oh right okay. understood. thanks