-1

I know that Z4×Z6 is non cyclic group. Does that follows it's quotient group is also non cyclic? I know that | (Z4×Z6) |/| (<2,2>)| is 4. The question was asked that what group it's isomorphic to, I wrote reasoning that it's isomorphic to Z2×Z2, and not isomorphic to cyclic group Z4 since Z4 is cyclic and the quotient group can not be cyclic because the group Z4×Z6 is also not cyclic. My question is I know that if quotient group is cyclic then group is also cyclic. But can I also make and argument that group is not cyclic so quotient group is also not cyclic?

user1100367
  • 1
  • 1
  • 1
    The claim that the quotient group cannot be cyclic because the original group is not cyclic is patently absurd. – Arturo Magidin Dec 03 '23 at 03:29
  • Since it is not in the answer linked by Moishe (at least not readily to my eyes) note that it is not the case that you can conclude $G$ is cyclic if you know $G/K$ is cyclic for some quotient $G/K$ of $G$. The simplest counterexample is $G=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ and $K$ any subgroup of order $2$. – krm2233 Dec 03 '23 at 03:31

1 Answers1

0

The answers in the linked question seem like overkill: if $G = \mathbb Z/4\mathbb Z \times \mathbb Z/6\mathbb Z$ and $H=\langle (2,2)\rangle$ then $$ H = \{(2,2),(0,4),(2,0),(0,2),(2,4),(0,0)\} $$ so that $G/H$ has order $24/6=4$. Now since $H$ contains every element of $G$ of the form $(2k,2l)$ it is clear that $(g+H)+(g+H) = 0+H$ and so every element of $G/H$ has order $2$, and hence $G/H$ cannot be cyclic, and thus $G/H \cong (\mathbb Z/2\mathbb Z)^2$.

If you know the Chinese Remainder Theorem then $\mathbb Z/\mathbb 6\mathbb Z \cong \mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z$, where $1 \mapsto (1,1)$ and hence $2\mapsto (0,1)$ from which it also follows immediately that $G/H\cong (\mathbb Z/2\mathbb Z)^2$.

krm2233
  • 7,230