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Which of the following groups are isomorphic?

(a) $\mathbb{R}$ and $\mathbb{C}$

(b) $\mathbb{R}^*$ and $\mathbb{C}^*$

(c) $S_3\times \mathbb{Z}_4$ and $S_4$

(d) $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\mathbb{Z}_4$

Here option (d) is not correct because one is not cyclic and other one is cyclic. Also $\mathbb{R}$ and $\mathbb{C}$ are vector space isomorphic (over the field $\mathbb{Q}$). But i cannot conclude correct answer help me!

Rusty
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S. Pitchai Murugan
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1 Answers1

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(a) As you have observed, these are isomorphic as $\mathbb{Q}$-vector spaces. A fortiori their underlying additive groups are isomorphic.

(b) Count the solutions of $x^4=1$ in $\mathbb{R}^*$ resp. $\mathbb{C}^*$. (Alternatively, look at the map $x \mapsto x^2$ and check for surjectivity.)

(c) The abelianization of $S_3 \times \mathbb{Z}/4$ is $\mathbb{Z}/2 \times \mathbb{Z}/4$, but the abelianization of $S_4$ is $\mathbb{Z}/2$. (Alternatively, notice that there is no epimorphism $S_4 \to \mathbb{Z}/4$ since the kernel would contain the commutator subgroup $A_4$, but it would have only $4!/4=6$ elements.)

(d) You are right, $\mathbb{Z}/2 \times \mathbb{Z}/2$ is not cyclic.

Martin Brandenburg
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  • Do you know if the first isomorphism still holds without choice? – Tobias Kildetoft Sep 13 '14 at 13:26
  • I don't think that one needs full choice, but I am pretty sure that ZF is not enough. On the other hand, look at http://mathoverflow.net/questions/23202 – Martin Brandenburg Sep 13 '14 at 13:33
  • Right, full choice probably has no chance of being needed here. I guess a more precise question would be if it is consistent with ZF that they are not isomorphic (Asaf probably knows, but obviously I cannot ping him here). – Tobias Kildetoft Sep 13 '14 at 13:38
  • Yes, it is consistent (already mentioned by Asaf here: http://math.stackexchange.com/questions/302514 ) – Martin Brandenburg Sep 13 '14 at 14:26
  • For (c) can we see that $S_4$ has only one subgroup of index $2$, but $S_3\times \mathbb Z_4$ has two? – Potla Nov 30 '15 at 20:30