GATE Overflow Test Series | Discrete Mathematics | Test 3 | Question: 17

Question

Which of the following is NOT correct? (Mark all the appropriate choices)

• A. Any non single element group $G$ has at least $2$ subgroups.
• B. Every subgroup of a cyclic group is cyclic.
• C. A cyclic group is not abelian.
• D. Any subgroup of an abelian group is also abelian.

Answer 1

Any non-single element group $G$ always has $2$ trivial subgroups, $G$ and $\{e\}$.

$\textbf{B. Proof:}$
Let $G$ be a cyclic group generated by $'a'.$ By definition of cyclic group, every element  of $G$ has the form $a^{n}$ for some $n \in \mathbb{Z}^+$

Let $H$ be a subgroup of $G$

If $H=\{e\}$, then $H$ is a cyclic subgroup generated by $e$

Let $H \neq \{e\}$

Then as $H$ is a subgroup of $G,$ $a^{n} \in H$ for some $n \in \mathbb{Z}^+$

Let $m$ be the smallest positive integer such that $a^m \in H.$ We claim that $a^m$ generates $H.$

Consider an arbitrary element $b$ of $H$.

As $H$ is a subgroup of $G,b=a^{n}$ for some positive integer $n.$

By the division theorem, it is possible to find integers $q$ and $r$ such that $n=mq+r$ with $0\leq r \leq m$.

It follows that:

$a^{n}=a^{mq+r}=(a^{m})^{q}a^{r}$
and hence:

$a^{r}=(a^{m})^{-q}a^{n}$

Since $a^{m} \in H$ so is its inverse $(a^{m})^{-1}$

By closure property of group, so are all powers of its inverse.

Now $a^{n}$ and $(a^{m})^{-q}$ are both in $H$, thus so is their product $a^{r},$ again by closure property of group.

However $m$ was the smallest (strictly) positive integer such that $a^{m}\in H$ and $0\leq r < m$

Therefore it follows that:

$r=0.$

Therefore: $n=qm$
and:
$b=a^{n}=(a^{m})^{q}$

We conclude that any arbitrary element $b=a^{n}$ of $H$ is a power of $a^{m}.$

So, by definition, $H=(a^{m})$ is cyclic.

Hence, any subgroup of cyclic group is cyclic.
Reference: https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic

$\textbf{C. Proof:}$ Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G= g$ (every element in $G$ is some power of $g$.)

Let $a$ and $b$ be arbitrary elements in $G$.

Then there exists $n,m\in \mathbb{Z}$ such that $a=g^{n}$ and $b=g^{m}$.

It follows that
$ab=g^{n}g^{m}=g^{n+m}=g^{m}g^{n}=ba$.

Hence, we obtain $ab=ba$ for arbitrary $a,b\in G$.

Thus, $G$ is an abelian group.

$\textbf{D. Proof:}$ If $H$ is a subgroup of an abelian group, and $x,y \in H$ then $x,y \in G$ and hence
$$\boxed{xy=yx}$$
So, the correct answer is $(C)$.