Written Assignment No. 1

Example of a Solved Exercise

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 $g\ne e$ but $g*g=e$.

Proof. Let $2n$ 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 $g\ne e$ in $G$ one would have $g\ne g^{-1}$, i.e., $g$ and $g^{-1}$ would be different elements. So the set $G-\left\{e\right\}$ would be the disjoint union of two element subsets of the form $\left\{g,g^{-1}\right\}$, and, therefore, the number $\left|G-\left\{e\right\}\right|$ of elements of $G-\left\{e\right\}$ would be even. Since $G$ is the disjoint union of $\left\{e\right\}$ and $G-\left\{e\right\}$, $$\left|G\right|=1+\left|G-\left\{e\right\}\right|,$$ 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 $G\left\{e\right\}$ for which $g*g=e$.

Assigned Exercises

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.

1. The multiplicative group of the integers mod $11$ is a cyclic group.

2. 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 $G$.