Cyclic Group Generators <z10, +> Mod 10 group of additive integers

[DUDEEGG]

So I take <z10, +> this to be the group


Z10 = {0,1,2,3,4,5,6,7,8,9} Mod 10 group of additive integers and I worked out the group generators, I won't do all of them but here's an example :

<3> gives {3,6,9,2,5,8,1,4,7,0}
on the other hand

<2> gives {2,4,6,8,0} and that's it! but my book says prove that 2 and 5 are generators...these two I thought we not generators...what am I doing wrong?

 

[jbunniii]

Are you sure that you are reading the question correctly? As you have noted, neither 2 nor 5 is a generator of $Z_{10}$. However, the set $\{2,5\}$ does generate $Z_{10}$.

 

[DUDEEGG]

Can you give me more info about {2,5} generating Z10? feel like I must be misunderstanding how to Generate Z10. Thanks for the help :)

 

[jbunniii]

We say that $\{2,5\}$ is a generating set for $Z_{10}$. Here is the definition:

http://en.wikipedia.org/wiki/Generating_set_of_a_group

Briefly, if $S$ is any subset of a group $G$, then we define $\langle S\rangle$ to be the smallest subgroup of $G$ which contains $S$. To be more precise, $\langle S\rangle$ is the intersection of all subgroups of $G$ which contain $S$. We call $\langle S\rangle$ the subgroup generated by $S$, and we say that $S$ is a generating set for $\langle S\rangle$.

In the special case where $S$ contains one element, say $S = \{x\}$, we usually write $\langle x \rangle$ instead of $\langle \{x\}\rangle$.

Applying these notions to your question, we want to prove that if $S = \{2,5\}$, then the smallest subgroup of $Z_{10}$ which contains $S$ is $Z_{10}$ itself. In other words, we want to prove that $\langle S\rangle = Z_{10}$.

To prove this, note that since $S$ contains $2$, it must be true that $\langle S\rangle$ contains $\langle 2\rangle$. (Why?) Likewise, since $S$ contains $5$, it must hold that $\langle S \rangle$ contains $\langle 5\rangle$. What can you conclude?

 

[HallsofIvy]

A group, G, is generated by a set, S, of members of G if and only if by combining the members of S using the group operation, we get every member of G. The set {3} gives, as you say all members of Z10 because 3= 3, 3+ 3= 6, 3+ 3+ 3= 9, 3+ 3+ 3+ 3= 12= 2 (mod 10), 3+ 3+ 3+ 3+ 3= 15= 5 (mod 10), 3+ 3+ 3+ 3+ 3+ 3= 18= 8 (mod 10), 3+ 3+ 3+ 3+ 3+ 3+ 3= 21= 1 (mod 10), 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3= 24= 4 (mod 10), 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3= 27= 7 (mod 10), and 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3= 30= 0 (mod 10).

{2} does not generate Z10 because 2= 2, 2+ 2= 4, 2+ 2+ 2= 6, 2+ 2+ 2+ 2= 8, 2+ 2+ 2+2+ 2= 10= 0 (mod 10), 2+ 2+ 2+ 2+ 2+ 2= 12= 2 (mod 10) and now it starts over. We do NOT get 1, 3, 5, 7, or 9.

{5} does not generate Z10 because 5= 5, 5+ 5= 10= 0 (mod10), 5+ 5+5 = 15= 5 (mod 10) and it just repeats. We do not get 1, 2, 3 4, 6, 7, 8, or 9.

But {2, 5} contains all of those and 2+ 5= 7, 2+ 2+ 5= 9, 2+ 2+ 2+ 5= 11= 1 (mod 10), 2+ 2+ 2+ 2+ 5= 13= 3 (mod 10) so that we have all of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0.

 

[DUDEEGG]

HallsofIvy EGGGZELLENT Thx :) u2
jbunniii