Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)

The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$. ...... (2)

If my statement is $p\to q$ then converse is $q\to p$.

i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $p\to q$ then converse is $q\to p$

  • $\begingroup$ (I'm sorry if this question is uninteresting But please help!) $\endgroup$ – Cloud JR Aug 12 '18 at 8:59
  • $\begingroup$ See math.stackexchange.com/a/41762/8581 $\endgroup$ – mrs Aug 12 '18 at 9:03
  • $\begingroup$ I don't want counter example! i want to know how (2) is converse of (1) $\endgroup$ – Cloud JR Aug 12 '18 at 9:05
  • $\begingroup$ I have given an answer for how to formulate the converse of the Lagrange's theorem. However, the other answers as for the counterexample to disprove the converse statement are also correct! $\endgroup$ – Aniruddha Deshmukh Aug 12 '18 at 9:05

Actually, you need to write the Lagrange theorem in $p \rightarrow q$ form. So, the Lagrange theorem is actually,

If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.

Here $p:$ $H$ is a subgroup of $G$

$q;$ Order of $H$ divides order of $G$.

Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.

So, the converse will be

If a number $m$ divides order of $G$, then there is a subgroup of order $m$.

I hope this gives you insights.

  • $\begingroup$ It seems that the OP gets trouble in understanding a logical statement. $\endgroup$ – mrs Aug 12 '18 at 9:07
  • $\begingroup$ @ResidentDementor yes you are correct $\endgroup$ – Cloud JR Aug 12 '18 at 9:08
  • $\begingroup$ Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra". $\endgroup$ – Aniruddha Deshmukh Aug 12 '18 at 9:08
  • $\begingroup$ well i change the tags.. thanks for your suggestion $\endgroup$ – Cloud JR Aug 12 '18 at 9:10
  • $\begingroup$ So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct? $\endgroup$ – Cloud JR Aug 12 '18 at 9:13

What is exactly your question? Do you want a counter example of the converse of Lagrange's theorem?

Such a counter example is follwing: The alternating group $A4$ has 12 elements, but doesn't contain a subgroup of order 6.

  • $\begingroup$ Better,you made this as comment $\endgroup$ – mrs Aug 12 '18 at 9:02
  • $\begingroup$ You are of course right. I'd like write this as a comment, but I don't have 50 reputation to do so. $\endgroup$ – mathlettuce Aug 12 '18 at 9:05
  • $\begingroup$ I don't want counter example! i want to know how (2) is converse of (1) $\endgroup$ – Cloud JR Aug 12 '18 at 9:05

Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!

  • $\begingroup$ I don't want counter example! i want to know how (2) is converse of (1) $\endgroup$ – Cloud JR Aug 12 '18 at 9:05
  • 2
    $\begingroup$ The order of $A_4$ is $12$ $\endgroup$ – leibnewtz Aug 12 '18 at 9:20
  • $\begingroup$ Yes, you're right @leibnewtz. My mistake !! $\endgroup$ – Anik Bhowmick Aug 12 '18 at 9:21

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