Group Homomorphism in Z_7 - Why is the Answer Yes?

[PsychonautQQ]

Homework Statement

Groups G and H are both groups in Z_7 (integer modulo), the mapping Is given by ø(g) = 2g

is ø: G-->H a homomorphism?

The Attempt at a Solution

My textbook says yes, I can't understand why.

ø(g1g2) = 2(g1g2) does not equal 2g1*2g2 = ø(g1)ø(g2)

something missing here?

 

[HallsofIvy]

Apparently you are missing that this is in "Z_7". (I'm not sure what you mean by "groups in Z_7". The group Z_7, with either addition as operation, has only itself and the identity as subgroups.)

In any case, in Z_7, ø(0)= 0, ø(1)= 2, ø(2)= 4, ø(3)= 6, ø(4)= 1, ø(5)= 3, ø(6)= 5.

I won't go through all 36 possible combinations but, for example, if g1= 3 and g2= 5, then 3+ 5= 8= 1 (mod 7) so ø(g1g2)= ø(1)= 2 while ø(3)= 6 and ø(5)= 3 and 6+ 3= 9= 2 (mod 7).

 

[PsychonautQQ]

Ooh thank you, I thought the group operation was multiplication

 

[jbunniii]

Ooh thank you, I thought the group operation was multiplication

$Z_7$ is the ring of integers modulo 7. Every ring is an additive group (just ignore the multiplication).

A ring is not a multiplicative group because $0$ has no multiplicative inverse, and the same may be true of other elements. However, the set of units (elements which have multiplicative inverses), if it is nonempty, does form a group under multiplication, called the group of units. To distinguish this group from the ring, we often use a notation such as $Z_7^\times$.

In the case of $Z_7$, or more generally $Z_p$ for any prime number, every nonzero element is a unit. So $Z_7^\times$ consists of $\{1,2,3,4,5,6\}$ and more generally, $Z_p^\times$ contains $p-1$ elements when $p$ is prime.

On the other hand, if $n$ is not prime, then some nonzero elements of $Z_n$ may not be units. For example, in $Z_4$, the group of units is $Z_4^\times = \{1,3\}$.