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Please note that the directions for this solved exercise differ from those for the exercises in the present assignment.
Prove the following statement: If the number of elements in a finite group G with identity e is even, show that there is at least one element g in G such that g ≠ e but g * g = e.
Proof. Let 2 n be the number of elements of the given finite group G. The assertion is that there is at least one element of G other than e for which g * g = e, i.e., g = g^{-1}. If this were not the case then for every g ≠ e in G one would have g ≠ g^{-1}, i.e., g and g^{-1} would be different elements. So the set G -
| { | e | } |
would be the disjoint union of two element subsets of the form
| { | g, g^{-1} | } |
, and, therefore, the number |G -
| { | e | } |
| of elements of G -
| { | e | } |
would be even. Since G is the disjoint union of
| { | e | } |
and G -
| { | e | } |
,
| |G| = 1 + |G - |
| | , |
and, therefore, the number of elements of G would be odd. Hence, if the number of elements of G is even, there must be at least one element of G
| { | e | } |
for which g * g = e.
Read these directions carefully: for each of the following statements either provide a proof that the statement is true or label the statement as false and provide justification.
The multiplicative group of the integers mod 11 is a cyclic group.
If G is an abelian group with identity e, then the set T of all elements t \in G such that t^{2} = e is a subgroup of G.