Question

1. Suppose G is an abelian group of order 6 containing an element of order 3. Prove G is cyclic?
2. Suppose G has only one element (say a) of order 2. show xa=ax for all x in G?

Answer 1

1. Suppose $G$ is an abelian group of order 6 containing an element $a$ of order 3. Let us prove $G$ is cyclic. By Cauchy's Lemma, the group $G$ contains an element $b$ of order 2. Since elements $a$ and $b$ are commute, and 2 and 3 are relatively prime, we conclude that the element $c=ab$ is of order $2\cdot 3=6$. Therefore, the cyclic subgroup $\langle c\rangle$ of $G$ has order 6. Since $G$ also has order 6, we conclude that $G=\langle c\rangle$. Therefore, $G$ is cyclic group generated by $c.$

2. Suppose $G$ has only one element (say $a$) of order 2. Let us show $xa=ax$ for all $x \in G$. Since for each $x\in G$, for the element $b_x=xax^{-1}$ we have that $b_x^2=b_xb_x=(xax^{-1})(xax^{-1})=xa(x^{-1}x)ax^{-1}=xa^2x^{-1}=xx^{-1}=e$, we conclude that for each $x\in G$ the element $b_x$ has order 2. Since $a$ is a unique element of order 2, we conclude that $b_x=a.$ Therefore, for each $x\in G$ we have that $xax^{-1}=a$, and hence each $x\in G$ we have that $xa=ax$.