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Let $G$ a group, $S,T \subseteq G$ and $\langle S \rangle, \langle T \rangle$ the subgroups generated by $S$ and $T$. Is it true or false that $$\langle S \rangle \cup \langle T \rangle = \langle S \cup T \rangle$$ and $$\langle S \rangle \cap \langle T \rangle = \langle S \cap T \rangle?$$

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  • $\begingroup$ Any thoughts? Any guesses? $\endgroup$ – Omnomnomnom Mar 5 '14 at 1:38
  • $\begingroup$ This can be shown false by very small counterexamples. Try $S_3$. $\endgroup$ – Vladhagen Mar 5 '14 at 1:38
  • $\begingroup$ @Omnomnomnom No $\endgroup$ – Twink Mar 5 '14 at 1:39
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    $\begingroup$ UNION DOSEN'T PLAY WELL WITH GROUPS! $\endgroup$ – RougeSegwayUser Mar 5 '14 at 1:59
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Let $S=(12)$ and let $T = (13)$. Then $\langle (12)\rangle \cup\langle (13)\rangle$ has three elements (And is not even a group!).

But, $\langle (12), (13)\rangle = S_3$. This has 6 elements and IS a group.

You can do something quite similar for the intersection. The elements $S$ and $T$ should actually work as a counter example..

Explicitly: $\{(12)\}\cap\{(13)\} = \emptyset$ as a set intersection. But the intersection of $\langle (12)\rangle$ and $\langle (13)\rangle$ contains the identity.

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  • $\begingroup$ But @Omnomnomnom said the second is true $\endgroup$ – Twink Mar 5 '14 at 1:43
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    $\begingroup$ And he/she is wrong. Hence the deleted answer! $\endgroup$ – Vladhagen Mar 5 '14 at 1:43
  • $\begingroup$ Bu the union is true right? Could you please write down a proof? Thank you in advance:) $\endgroup$ – user557550 Mar 20 at 1:07
  • $\begingroup$ The statement on the union of two groups is also false. I have shown a counter example in my answer. $\endgroup$ – Vladhagen Mar 20 at 2:20
  • $\begingroup$ And $\langle \langle S \rangle \cup \langle T \rangle \rangle = \langle S \cup T \rangle$? $\endgroup$ – user557550 Mar 21 at 15:59

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