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Lagrange: Let $G$ a finite group and $H$ a subgroup of $G$. Then the order of $H$ divides the order of $G$.

I know that converse of Lagrange Theorem is not necessarily true. But my teacher gave me an exercise to determine if given a group $G$ of order 6, then $G$ has subgroups of orders $1, 2, 3$ and $6$. I also know, there are $2$ group structures up to isomorphism ($S_3$ and $\Bbb{Z}/6\mathbb{Z}$) and both of them have subgroups of order $1, 2, 3$ and $6$. Can I conclude that given a group or order $6$, it has subgroups of order mentioned before by that statement?

Shaun
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Jhon C
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    Yes. The smallest counterexample to the converse of Lagrange's Theorem has order $12$. – Arturo Magidin Jul 10 '22 at 17:53
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    It's not "reciprocal", it's "converse" – Arturo Magidin Jul 10 '22 at 17:54
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    You say yourself that there are only two cases, and that you can treat each of them. What could be the issue? – Captain Lama Jul 10 '22 at 17:56
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    See also here for more information on CLT groups (Lagrangian groups). – Dietrich Burde Jul 10 '22 at 18:07
  • @CaptainLama I am a beginner with group theory and the concept of isomorphism is new for me. If two groups are isomorphic I understand they are similar, they share similar characteristics (order, cyclics, abelians,...) but I thought, maybe, the number of subgroups with the same order would not be one of those characteristics. – Jhon C Jul 10 '22 at 18:10
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    When two groups are isomorphic any true statement about one of them that depends only on the group axioms is true for the other. They are "the same groups" with different names for the elements. See https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 – Ethan Bolker Jul 10 '22 at 18:26

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So, the result relies on the knowledge that there are, up to isomorphism, only two groups of order six.

We know darn well that $S_3$ and $\Bbb Z_6$ are CLT groups.

By the structure theorem, the only abelian group of order six is the cyclic one.

It would take some doing, (just as the structure theorem takes doing), but $S_3$ is the only nonabelian group of order six.

So we really are standing on the shoulders of giants here.

Finally, as @Arturo noted, the smallest non CLT group is $A_4$, which has no subgroup of index $2$.

calc ll
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