Question

Show that for every group with an even number of elements that at least two elements must be their own inverse.

Answer 1

@ikram002p

Answer 2

assume group G
G={a1,a2,....a2n} of order 2n s.t n in Z
so
\[a_i \neq a_j \forall i,j \in Z\]

Answer 3

\[i \neq j /also\]

Answer 4

\[a_i \neq a_j\]
let b is the inverse for both ai & aj then
\[ba_i \neq ba_j..........eq1\]
\[ba_i =e\]
\[ba_j =e\]
from eq1
\[e\neq e\] wich is a contradiction
so inverse of a_i must be different than inverse of a_j

Answer 5

wait but how do we go from a1...an to ai and aj?

Answer 6

simply
i,j <=n
1 belong to n
its just numbers of order

Answer 7

if u wanna use sempols
G={a,b,c,d,.....,z} of order 2n (even number )

Answer 8

ok thank youu

Answer 9

\[a \neq b \neq c \neq d \neq z....\]
its the same for
\[a_i\neq a_j \forall i,j \in Z s.t ....i \neq j \]

Answer 10

np ur wlc anytime :)