1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove that no group of order 160 is simple

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that no group of order 160 is simple.

    2. Relevant equations
    Sylow Theorems, Cauchy's Theorem, Lagrange's Theorem.


    3. The attempt at a solution
    Because [itex]160 = 2^5×5[/itex], by the First Sylow theorem, there is a subgroup [itex] H [/itex] of order [itex]2^5 = 32[/itex] in [itex]G[/itex]. Let [itex] S [/itex] be the set of all left-cosets of [itex]H[/itex] (as of now, it may not be a group). By Lagrange's Theorem, [itex]|S| = [G:H] = |G|/|H| = 5[/itex]. Consider the set [itex]S' = \{H, gH, g^2H, g^3H, g^4H\} [/itex] where [itex] g [/itex] has order 5 (by Cauchy's Theorem there exists a subgroup of order 5 in [itex]G[/itex]). [itex]S' \subseteq S[/itex] because it consists of (not necessarily distinct) left-cosets of [itex]H[/itex]. But suppose [itex]g^iH = g^jH[/itex] for some [itex] 0≤i, j≤4 [/itex]. Then by basic theorems of cosets, [itex] g^i * (g^j)^{-1} = g^{i-j} \in H [/itex]. But [itex]g[/itex] has order a power of 5, and [itex]H[/itex] only contains elements of order power of 2, so [itex]g^{i-j} = e [/itex] and [itex]g^i = g^j[/itex]. This proves that all elements of [itex]S'[/itex] are distinct, and since [itex] |S| = |S'| = 5 [/itex] and [itex] S' \subseteq S[/itex], [itex]S = S'[/itex].

    We have proven that the set of all cosets of [itex]H[/itex] is [itex]S = \{H, gH, g^2H, g^3H, g^4H\} [/itex]. But this set forms a group under coset multiplication, as can be verified from the axioms for a group. Here the less obvious part is to show the multiplication is well-defined. But every coset can be written in a unique way as [itex]g^iH, 0≤i≤4[/itex] so the result of the multiplication [itex]g^iH * g^jH = g^{i+j}H[/itex] is always well-defined. The operation obviously respects closure, since [itex]g^{i+j}H[/itex] is a coset of [itex]H[/itex]. The identity and inverses are also easy to find, and associativity follows from associativity of addition in [itex]\mathbb{Z}_5[/itex].

    From this, it follows that the set of left-cosets of [itex]H[/itex] forms a group under coset multiplication, and it is the quotient group [itex]G/H[/itex]. But quotient groups are defined if and only if [itex]H[/itex] is a normal subgroup, which proves [itex] H [/itex] is a nontrivial, proper normal subgroup of [itex]G[/itex]. Therefore, [itex]G[/itex] is not simple.
     
    Last edited: Aug 25, 2013
  2. jcsd
  3. Aug 25, 2013 #2
    I can't find any mistakes after re-reading my proof, but for some reason the proof looks fishy to me. The reason is that I prove a subgroup H is normal in G by actually proving the quotient group G/H exists. Usually, we first prove that a group is normal, and then we define the quotient group. It may work, but I would like to have some confirmation in case I overlooked something. When I think about it, it looks even more suspicious because the same argument would show that every subgroup that has a prime number of cosets is a normal subgroup. I don't know if that is true.
     
    Last edited: Aug 25, 2013
  4. Aug 26, 2013 #3
    Ah I think I found where the problem is!
    My mistake was to define the coset multiplication only for one specific coset representative. It is true that [itex]g^iH * g^jH = g^{i+j}H[/itex], but if I chose different coset representatives, [itex]aH = g^iH, bH = g^jH[/itex] then [itex]aH * bH[/itex] is not necessarily [itex](ab)H[/itex] because [itex]ab[/itex] might not be another coset representative of [itex]g^{i+j}H[/itex].

    In this case I admit I am lost on this question. Also, sorry for the multiple posts, but I can't edit the previous posts anymore.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted