How do i prove that any subgroup of $A_5$ has order at most 12? [duplicate]

I know this question has been answered Other proofs that subgroups of $$A_5$$ have order at most 12

But i have difficulty in understanding that proof.The book says we can assume that $$A_5$$ has no normal subgroup.How to find the proof using this property ?

EDIT- (I solved the previous question ( $$A_5$$ has no normal subgroup ) but i am not able to solve this problem ).

I have already mentioned that my question similar to Other proofs that subgroups of $$A_5$$ have order at most 12 so please don't mark it as duplicate.

The book which i am talking about is - Topics In Algerbra by Herstein. (2.10.15)

I am thankful if some can explain the same proof ( provided in link ).I have difficulty in understanding the homomorphic part of that answer.

• @JoséCarlosSantos I have mentioned it in my question .I have difficulty in understanding that proof.I want to prove it using the fact that $A_5$ has no normal subgroup . – user614560 Nov 12 '18 at 12:31
• Then you comment that answer. Or you reproduce the answer within your question explaining where did you get stuck. – José Carlos Santos Nov 12 '18 at 12:32
• @JoséCarlosSantos Any hint or suggestion from your side ? – user614560 Nov 12 '18 at 12:37
• @JoséCarlosSantos The user of that answer is probably not on stack exchange .Also i think that answer does not proves by using the property that $A_5$ has no normal subgroup – user614560 Nov 12 '18 at 12:43
• The homomorphic map, both in the lin and in my answer, refers to a homomorphism from the group $\;A_5\;$ to the group of permutation on $\;l\;$ elements, whith $\;l=\,$ the number of cosets of the subgroup. This is not a subgroup of $\;A_5\;$, only a set upon which $\;A_5\;$ acts... – DonAntonio Nov 12 '18 at 13:38

Let $$\;H\;$$ be a subgroup of $$\;A_5\;$$ of order $$\;>12\implies\;$$ its index is $$\;l<5\;$$ . Then $$\;A_5\;$$ acts on the set of left cosets of $$\;H\;$$ in $$\;G\;$$ , and this determines a homomorphism $$\;\phi:A_5\to S_l\;$$ . Since $$\;A_5\;$$ is simple this homomorphis is acutaly a monomorphism (i.e., $$\;1-1$$) , and thus it is an injection, which of course is impossible as $$\;|A_5|=60>S_l\;$$

• Index can be equal to 4 .Suppose subgroup is of order 15 – user614560 Nov 12 '18 at 13:38
• @Amit Of course, that was just a typo. Edited now, thanks. – DonAntonio Nov 12 '18 at 13:39
• This is a very beautiful proof and it can be extended to show that for every $n\ge 5$, $A_n$ has no subgroups of order larger than $\frac{(n-1)!}{2}$. – user593746 Nov 12 '18 at 13:43
• A simple group $G$ has only two normal subgroups$\text{---}$$\{e\}$ and the whole $G$. Since the kernel of any homomorphism of groups $\psi:G\to K$ is a normal subgroup of $G$, we have $\ker \psi=\{e\}$ or $=G$. So, either $\phi$ is injective or trivial. In the case at hand, $H$ is assumed to be a proper subgroup of $A_5$ (I forgot to add the word "proper"). So, $A_5$ cannot act on $A_5/H$ trivially. That is, $\phi:A_5\to S_l$ can't be the trivial homomorphism, so it's injective. – user593746 Nov 12 '18 at 14:01
• Thanks @DonAntonio .Even after being duplicate question you took out time and cleared my doubt.Math SE people should understand that there can be slow learner's . – user614560 Nov 12 '18 at 15:27

$$A_5$$ has order 60. Thus, any proper subgroup $$H$$ of order larger than 12 must have order 15, 20, or 30. Any subgroup of order 30 would have index 2, so would be normal, contradicting your previous result. Thus, $$H$$ must have order 15 or 20.

Now, both 15 and 20 are multiples of 5, so $$H$$ must contain an element of order 5. Up to conjugation, that element is $$(1,2,3,4,5)$$. If $$H$$ has order 15, then it also contains an element of order $$3$$, which must be a 3-cycle. But running through each possible 3-cycle:

$$(1,2,3)(1,2,3,4,5)^2 = (1,4)(2,5)$$, which has order 2, so $$H$$ has even order, a contradiction.
$$(1,2,4)(1,2,3,4,5)^4 = (2,3)(4,5)$$, so again, $$H$$ has even order. $$(1,3,5)(1,2,3,4,5)^4 = (1,2)(3,4)$$, so again, $$H$$ has even order.

And all others are conjugate (as a whole) to those, so symmetrically, have the same orders. Thus, there is no subgroup of order 15.

Thus, $$H$$, if it exists, must have order 20, so must contain an element of order 2, which must be a product of two disjoint 2-cycles. Again, running through the possibilities:

$$(1,2)(3,4)(1,2,3,4,5) = (1,3,5)$$, so $$H$$ has order 60.
$$(1,3)(2,4)(1,2,3,4,5)^2 = (1,5,2)$$, so $$H$$ has order 60.
$$(1,3)(2,5)(1,2,3,4,5)^2 = (1,5,4)$$, so $$H$$ has order 60.
$$(1,4)(2,5)(1,2,3,4,5) = (3,5,4)$$, so $$H$$ has order 60.
$$(1,5)(2,3)(1,2,3,4,5) = (2,4,5)$$, so $$H$$ has order 60.
$$(1,5)(3,4)(1,2,3,4,5) = (2,3,5)$$, so $$H$$ has order 60.
$$(2,3)(4,5)(1,2,3,4,5) = (1,2,4)$$, so $$H$$ has order 60.
$$(2,4)(3,5)(1,2,3,4,5)^2 = (1,3,2)$$, so $$H$$ has order 60.

Now, you'll notice that I've missed a few permutations, namely:

$$(1,2)(3,5)$$
$$(1,3)(4,5)$$
$$(1,4)(2,3)$$
$$(1,5)(2,4)$$
$$(2,5)(3,4)$$

With these, we have:
$$(1,2)(3,5)(1,2,3,4,5) = (1,3)(4,5)$$,
$$(1,3)(4,5)(1,2,3,4,5) = (1,4)(2,3)$$,
$$(1,4)(2,3)(1,2,3,4,5) = (1,5)(2,4)$$,
$$(1,5)(2,4)(1,2,3,4,5) = (2,5)(3,4)$$, and $$(2,5)(3,4)(1,2,3,4,5) = (1,2)(3,5)$$,

so if any one of these lies in $$H$$, all of them do. Further, these five elements, along with the powers of $$(1,2,3,4,5)$$ form a subgroup, of order 10. But if $$H$$ contains any other element, it must be one of order 2 or 5 (else $$H$$ would have order a multiple of 3): the order 2 cases have been covered above, and all give $$H = A_5$$. Checking the elements of order 5 that aren't already in there:

$$(1,2,3,4,5)(1,2,3,5,4) = (1,3)(2,5)$$,
$$(1,2,3,4,5)(1,2,4,3,5) = (1,4)(2,5)$$, $$(1,2,3,4,5)(1,2,4,5,3)^2 = (1,5,4)$$, $$(1,2,3,4,5)(1,2,5,3,4)^2 = (1,3,2)$$, $$(1,2,3,4,5)(1,2,5,4,3) = (1,5,2)$$, $$(1,2,3,4,5)(1,3,2,4,5) = (1,4)(3,5)$$, $$(1,2,3,4,5)(1,3,2,5,4) = (1,5,3)$$,
$$(1,2,3,4,5)(1,3,4,2,5)^2 = (3,5,4)$$,
$$(1,2,3,4,5)(1,3,4,5,2) = (2,4)(3,5)$$,
$$(1,2,3,4,5)(1,3,5,4,2) = (2,5,3)$$,
$$(1,2,3,4,5)(1,4,2,3,5)^2 = (1,5,2)$$,
$$(1,2,3,4,5)(1,4,3,2,5) = (1,5,4)$$,
$$(1,2,3,4,5)(1,4,3,5,2) = (2,5,4)$$,
$$(1,2,3,4,5)(1,4,5,2,3)^2 = (2,4,3)$$,
$$(1,2,3,4,5)(1,4,5,3,2) = (3,5,4)$$,
$$(1,2,3,4,5)(1,5,2,3,4) = (1,3)(2,4)$$,
$$(1,2,3,4,5)(1,5,2,4,3) = (1,4,2)$$,
$$(1,2,3,4,5)(1,5,3,2,4) = (1,4,3)$$,
$$(1,2,3,4,5)(1,5,3,4,2) = (2,4,3)$$,
$$(1,2,3,4,5)(1,5,4,2,3) = (1,3,2)$$.

So if $$H$$ contains any of these, it is all of $$A_5$$. Thus, there is no subgroup of $$A_5$$ of order 20, and hence no proper subgroup of order more than 12.

[NB: there are some pretty massive time-saving options available here: I went for keeping things as elementary as possible]

• It might help if you use the fact that any group of order $15$ is cyclic. But there is no element of $A_5$ of order $15$. – user593746 Nov 12 '18 at 14:39
• Indeed, that's one of the time-saving options that I mentioned. – user3482749 Nov 12 '18 at 14:54