- #1
Boorglar
- 210
- 10
Homework Statement
Prove that no group of order 160 is simple.
Homework Equations
Sylow Theorems, Cauchy's Theorem, Lagrange's Theorem.
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] = |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: