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I know that $A_4$ group of order $12$ does not have the subgroup of order $6$. I know how to prove that .
My teacher says that it is the smallest such example.
Can we prove this fact with rigorous proof? Or do we have to give case by case like $4$ order, $5$ order ... group until we hit $12$?
Any Help will be appreciated.

ArsenBerk
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Curious student
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    You seem to be implying that simply checking that the result does not hold for the 19 groups of order less than twelve isn't "rigorous"?! – user1729 Sep 19 '18 at 08:35

2 Answers2

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To an extent you have to consider all numbers $\le11$. But if you have some knowledge of group theory, you can dispose of many of them quickly.

If you know Cauchy's theorem to the effect that if $p$ is a prime divisor of $|G|$ then $G$ has an element of order $p$, then one easily does the case where $|G|=pq$, with $p$, $q$ distinct primes.

Another useful theorem is that a group of order a prime power has subgroups of order every factor of $|G|$.

With both of these results at hand, then the theorem is already proved...

Angina Seng
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  • This indeed reduces to proving that all groups of order $8$ have a subgroup of order $4$. There are five of them: https://groupprops.subwiki.org/wiki/Groups_of_order_8 – egreg Sep 19 '18 at 09:09
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There are several results on so called Lagrangian groups, i.e., on groups where for each positive divisor $d$ of $|G|$ there exists at least one subgroup $H\le G$ with $|H|=d$, see here:

Complete classification of the groups for which converse of Lagrange's Theorem holds

For example, Lagrangian groups are supersolvable, but $A_4$ is not. And indeed, the smallest example of a finite non-supersolvable group is $A_4$.

Dietrich Burde
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