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Homework Help: 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.
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