To prove this is a group it will require us to prove associativity of DP_{(5,5)} as well as DP_{(5,5)} containing identity and the inverses

1. Associativity

Let

(2)
\begin{align} (a,b),(c,d),(e,f)\in DP_{5,5} \end{align}

we need to show

[(a,b)+(c,d)]+(e,f)= (a,b)+[(c,d)+(e,f)]

so simplifying

(a+c,b+d)+(e,f)=(a,b)+(c+e,d+f)

(a+c+e,c+d+f)=(a+c+e,c+d+f)

so

[(a,b)+(c,d)]+(e,f)= (a,b)+[(c,d)+(e,f)] is true

2. Identity

The identity of Z_{5} is 0 so the identity of DP_{(5,5)} is (0,0)

This is true since

(0,0)+(a,b)= (a,b)

3. inverses

inverses are also in DP_{(5,5)}

if we let a be an element of Z_{5} then there is an element in Z_{5} we will call a' that takes a to the identity.

we know this because we can lay out all the possible inverses of every element in Z_{5}.

a a'

0 0

1 4

2 3

3 2

4 1

so if this is the case then there is an (a,b) and (a',b') in DP_{(5,5)} such that

(a,b)+(a',b')=(0,0)

because we know that (a,b)+(a',b')= (a+a',b+b')=(0,0)

Thus we have proven DP_{5} is a group