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 $S' = \{H, gH, g^2H, g^3H, g^4H\}$ where $g$ has order 5 (by Cauchy's Theorem there exists a subgroup of order 5 in $G$). $S' \subseteq S$ 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 $S'$ are distinct, and since $|S| = |S'| = 5$ and $S' \subseteq S$, $S = S'$.

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.