Prove that no group of order 160 is simple

[Boorglar]

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 160 = 2^5×5, by the First Sylow theorem, there is a subgroup H of order 2^5 = 32 in G. Let S be the set of all left-cosets of H (as of now, it may not be a group). By Lagrange's Theorem, |S| = [G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />] = |G|/|H| = 5. Consider the set S&#039; = \{H, gH, g^2H, g^3H, g^4H\} where g has order 5 (by Cauchy's Theorem there exists a subgroup of order 5 in G). S&#039; \subseteq S because it consists of (not necessarily distinct) left-cosets of H. But suppose g^iH = g^jH for some 0≤i, j≤4. Then by basic theorems of cosets, g^i * (g^j)^{-1} = g^{i-j} \in H. But g has order a power of 5, and H only contains elements of order power of 2, so g^{i-j} = e and g^i = g^j. This proves that all elements of S&#039; are distinct, and since |S| = |S&#039;| = 5 and S&#039; \subseteq S, S = S&#039;.

We have proven that the set of all cosets of H is S = \{H, gH, g^2H, g^3H, g^4H\}. 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 g^iH, 0≤i≤4 so the result of the multiplication g^iH * g^jH = g^{i+j}H is always well-defined. The operation obviously respects closure, since g^{i+j}H is a coset of H. The identity and inverses are also easy to find, and associativity follows from associativity of addition in \mathbb{Z}_5.

From this, it follows that the set of left-cosets of H forms a group under coset multiplication, and it is the quotient group G/H. But quotient groups are defined if and only if H is a normal subgroup, which proves H is a nontrivial, proper normal subgroup of G. Therefore, G is not simple.

 

[Boorglar]

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.

 

[Boorglar]

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 g^iH * g^jH = g^{i+j}H, but if I chose different coset representatives, aH = g^iH, bH = g^jH then aH * bH is not necessarily (ab)H because ab might not be another coset representative of g^{i+j}H.

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.