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Let $G$ be a group with order 12. Which of the following claims are false?

Does Lagrange's Theorem imply that there could exist a subgroup with order 6? I'm not sure where to begin. The question was taken from a past test with multiple choice answers, and the question asked which of the following claims were false:

• G must have an element of order 2.
• G must have a subgroup of order 6.
• G must have a subgroup of order 2.
• None of the above, they are all true.

HoopsMcCann
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    How is it a "solution" if it does not justify its claim (when the only problem here that needs solving is why the claim is true). Also no, Lagrange theorem only tells you that $G$ cannot have a subgroup of order $5$, $7, 8, \ldots, 11$. – M. Vinay Mar 19 '19 at 02:14
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    Several groups of order 12 have subgroups of order 6. Is the problem translated incorrectly? – Randall Mar 19 '19 at 02:15
  • You must mean that "$G$ may not have a group of order $6$". Here, the "may not" is not imperative — it does not mean "must not". It means, "It is not necessary that $G$ has a subgroup of order $6$". – M. Vinay Mar 19 '19 at 02:16
  • Sorry for the confusion, I made an edit to the original post. – HoopsMcCann Mar 19 '19 at 02:17
  • The statement that the statement "$X$ must be true" is false does not say that "$X$ must be false". It means "$X$ may (or may not) be true". – M. Vinay Mar 19 '19 at 02:18
  • Now, as for the given statements, observe first of all that the first and third statements are equivalent (think about this and make sure you see it). So either the second statement is true, or, as the fourth statement is correct and all the above statements are true. (That's an odd option though, because if you select it, you'd be stating that it's false, since you're supposed to select the false statement. But if it is indeed false, then one of the above statements must be false, so that should be your answer, which is a contradiction….) – M. Vinay Mar 19 '19 at 02:22
  • But "it must not" is inherently ambiguous, and could have two equally sensible interpretations. The standard English interpreation is different from the German one, for example. I will edit the question to clarify it. – Derek Holt Mar 19 '19 at 02:23
  • @M.Vinay I don't quite see why the first and third statements are equivalent. Does it have to do with cyclic groups? – HoopsMcCann Mar 19 '19 at 02:29
  • @user2965071 Yes. If an element has order $2$, what is the order of the cyclic subgroup it generates (which is necessarily a subgroup of the group)? On the other hand if a (sub)group has order $2$, what must be the order of its elements? – M. Vinay Mar 19 '19 at 02:37
  • Consider a dihedral group of order $6$ I.e., $D_6$. –  Mar 19 '19 at 02:53

1 Answers1

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The first option is true because of Cauchy's theorem. This will also imply the existence of subgroup of order $2$(consider the subgroup generated by element of order $2$). Second option is false because $A_4$ is a group of order $12$ which has no subgroup of order $6$.

cqfd
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