Cyclic group

**chiph588@** · March 16th 2009, 12:54 PM

Let $G$ denote a group and let $g \in G$ . In each case show $G = < g >$ .

a. $|G| = 12, g^4 \neq 1, g^6 \neq 1$ .
b. $|G| = 40, g^8 \neq 1, g^{20} \neq 1$ .
c. $|G| = 12, g^{30} \neq 1, g^{20} \neq 1, g^{12} \neq 1$ .
d. Generalize.

**Opalg** · March 16th 2009, 02:14 PM

Originally Posted by chiph588@

Let $G$ denote a group and let $g \in G$ . In each case show $G = < g >$ .

a. $|G| = 12, g^4 \neq 1, g^6 \neq 1$ .
b. $|G| = 40, g^8 \neq 1, g^{20} \neq 1$ .
c. $|G| = 12{\color{red}0}, g^{30} \neq 1, g^{20} \neq 1, g^{12} \neq 1$ .
d. Generalize.

The order of an element of a group always divides the order of the group (so in example c., |G| can't be 12, it must be 120).

In example a., the order of g must be 1, 2, 3, 4, 6 or 12 (those being the divisors of 12. If g has order 1 then g is the identity and so g to any other power is also 1. If g has order 2 then $g^4 = 1$ . If g has order 3 then $g^6=1$ . So if $g^4\ne1$ and $g^6\ne1$ then the only remaining possibility is that g has order 12. Thus g generates the whole of G.

The other two examples are similar. I'll leave you to think about the generalisation.

Thread: Cyclic group

LinkBack

Thread Tools

Display

Cyclic group

Similar Math Help Forum Discussions

Order of Group. Direct Product of Cyclic Group

cyclic group

Cyclic Group

Prove cyclic subroups => cyclic group

automorphism group of a cyclic group

Search Tags