[8]=[2]ˉ¹ ?

• May 13th 2009, 08:44 PM
yvonnehr
[8]=[2]ˉ¹ ?
Why is this true? I don't understand.

Example 2 p164
It states that in the subgroup H={[2], [4], [6], [8]},
[8]=[2]ˉ¹ in Z₁₀ under multiplication.

This is what I gather:
Since [2] ∈ H, [8]=[2]=[2]ˉ¹.

or

Since [8] can be written as [8]³=[12]=[2], then [2]ˉ¹ (the inverse of [2]) also equals [8].

Please let me know if this is right and what definition or theorems support this. I think I heard something like this in class but I can't remember.(Doh)
• May 13th 2009, 09:29 PM
Jose27
I don't really understand what you are trying to do, but here's what I think you mean: That in the ring of integers modulo 10, 2*8=1 mod(10), but this is not true since the inverse of 2 is 5, and 2*8=16=6 mod(10). Also, I don't know what H has to do here, since operations on H are inherited from the whole ring.

Now if you mean that 2+8=0 mod(10) then this is indeed true because 10 divides 10, and hence 2 is the inverse of 8 in the ADDITIVE subgroup H.
• May 14th 2009, 12:57 AM
Swlabr
Hint: What is the identity of your group?

Further hint: \$\displaystyle (Z_{10}, *)\$ does not form a group, and neither does \$\displaystyle (Z_{10} \setminus \{0\}, *)\$ because in one you have a zero, and in the other \$\displaystyle 5*2=0\$.