No subgroup of $D_3$ has order 4.

Question

Question: Prove that the dihedral group of order $6$ has no subgroup of order $4$.

I am trying to prove above question. But i have no idea how can i start! Just before this question (in exercise) i prove a question, which is following:

A group with two elements of order $2$ that commute must have a subgroup of order $4$.

Now i am thinking that the above proposition will be used to prove that main question.

My attempt: Assume to the contrary that $D_3$ has a subgroup $H$ of order $4$. In view of previous question (above proposition) $D_3$ must satisfy the assumptions of proposition. Let $r$ and $f$ be rotation and reflection respectively in $D_3$. Since only reflections, $f,rf,r^2f$, have order $2$ in $D_3$ so any two of them must commute. But reflections don't commute with each other. Thus this contradicts that $D_3$ has a subgroup of order $4$.

My confusion: I am thinking my proof is incorrect because the proposition says "If there exists $a,b ∈G$ s.t $|a|=|b|=2$ and $ab=ba$, then there exists $H$ a subgroup of $G$ s.t $|H|=4$". But i am using the converse of the statement without knowing it is either true or false.
And another question if the proof is correct, then is it written rigorously? (I have always in my proof writing skills).
Note : I am trying to prove the question without Lagrange's theorem.
Thank you.

Answer 1

Lagrange's theorem would make the answer easy ($4$ does not divide $6$),

but consider that $D_3$ has elements of order $1, 2,$ and $3$.

The subgroup that contains only an element of order $2$ (besides the identity element) has order $2$.

The subgroup that contains only the elements of order $3$ (besides the identity element) has order $3$.

The subgroup that contains neither is trivial.

The subgroup that contains both is the entire group (order $6$).

Thus, there is no subgroup of order $4$.

Answer 2

Regarding the proof you have provided: this is not valid, unfortunately. Reason? You assume for the sake of contrapositive that there exists a subgroup order four in D3; this is fine. What's not fine is the assumption that in order for a group order four to exist in D3, the subgroup of interest must necessarily be generated by two elements order two which commute with each other. To reach this assumption would involve utilizing the converse of the proposition you provided which, recall, the following statement from that logic course you probably took first year of undergraduate: if it's outside raining, then it must be cloudy. Could we reasonably conclude that whenever it's cloudy, it must be raining? (answer: no).

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Seeing that you are not allowed to use Lagrange's Theorem, the solution your professor (or whoever) may be interested in seeing is one which may not be the most satisfying. As such, I present you the following list of things to look out for as to guide you to an appropriate proof:

(i) List all the elements as well as their orders. Could there exist a cyclic subgroup order four in D3?

(ii) Would there exist any contradictions with including an element order 3 in an order four subgroup? What about an element order 6?

(iii) Look now at the remaining elements. How many remain? Is it possible you could construct a subgroup order four using only these elements?