Prove that every even ordered group has an element of order 2.?

Question

Show that a group of order 12 has an element of order 3. Determine the order of 11 in the group U 237...

Answer 1

I can do the last 2.
Since 12 is divisible by 3, every group G of order 12
has a subgroup H of order 3 by Sylow's theorem.
But H is of prime order, so is cyclic. So H and
thus G contains an element of order 3.

U237 is the group of units mod 237.
We will use Euler's theorem:
If (a,n) = 1 then a^φ(n) = 1(mod n), where φ(n) is
Euler's phi function.
Next 237= 79*3,
so φ(237) = 78*2 =156.
So the order of 11(mod 237) must divide 156 = 2² * 3 * 13.
Now, with the aid of PARI, we get
11² = 121
11³ = 146
11^4 = 84
11^6 = 223
11^12 = 196
11^13 = 23
11^ 26 = 55
11^39 = 80
11 ^78 = 1
Thus the order of 11 in the group U237 is 78.

Answer 2

If every element generates a cyclic subgroup with an odd order, then the group will be of odd order, so there must be an element , call it a, which generates a cycle of length 2n for some interger n. Then a^n generates a subgroup of order 2.

Answer 3

(a) save in mind that for a p-group, its center is nontrivial (that's a results of the class equation). If |Z(G)| = p^2, then G = Z(G) so as that G is abelian. If |Z(G)| = p, then |G/Z(G)| = p, this ability that G/Z(G) is cyclic (of order p). that's at as quickly as forward to show that this implies that G is abelian. (actual that's reworded for further clarity for the naysayers available...) (b) a collection of order p^n is in many situations abelian whilst n <= 2. (n = a million grants a cyclic group, and n = 2 is worked out partly a.) on the various hand, if |G| = p^n with n >= 3, then there are constantly non-abelian communities to boot. i'm hoping that helps!