# Alternative way to prove that a group of order 160 is not simple

Is there a neat way to show that a group of order 160 is not simple without directly quoting Poincare's theorem?

I am thinking of maybe using the Sylow theorems to say that in order that the group is not simple we must have $n_2=5,\,\,\,n_5=2^4$ (due to the Sylow constraints). So then there are $2^4\times 4$ order 5 elements. But then I can't say that there are $5\times 2^5$ elements with not-equal-to-5 orders since these subgroups may intersect because they don't have prime orders. So what can we do next, or is this not even the right direction to start?

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Is there any reason why you would like to avoid Poincare? –  user21436 Jan 27 '12 at 21:40
@KannappanSampath: I have nothing against Poincare per se, but I was told that there is not-too-convoluted way to solve it... And maybe the alternate way would be illuminating in some way! ;) –  group Jan 27 '12 at 21:44
You can get the number of even-order elements up to 15 + 4*8, since each sylow-2 subgroup must at least be different from the others in the 8 elements with order relatively prime to the group order, but that still only gets you to a total of 112 elements accounted for. –  Carl Jan 27 '12 at 21:48
Counting arguments will go through. Look at the Normalizer of the intersection of two Sylow 2-Subgroups. (Also note that you cannot have more than one distinct Sylow 2-Subgroup.) –  user21436 Jan 27 '12 at 21:49
Another way I just thought of: Count the 2-Sylow-subgroups. There is either a single one (which consequently would be normal) or 5 distinct ones. In the later case the group acts transitively on 5 points and we get a non-trivial homomorphism into $S_5$ which is a group of order $120$. Therefore the kernel has order greater than $1$ and is a non-trivial normal subgroup. –  Sebastian Schoennenbeck Jan 27 '12 at 22:42

Let $H_1$ and $H_2$ be two Sylow $2$-Subgroups. Then, $|H_1|=|H_2|=32$. So, we see that $|H_1 \cap H_2| \le 16$.

Now Let the normalizer of their intersection be $N(H_1 \cap H_2)$. Notice that $|H_1 \cap H_2| \in \{2,4,8,16\}$. Now, I claim that it cannot be $2$ or $4$. To see this, look at $H_1H_2 \subset G$.

$$|H_1H_2|=\dfrac{|H_1||H_2|}{|H_1 \cap H_2|} \le 160 \implies |H_1 \cap H_2| \ge 7$$

Case 1:

Suppose $|H_1 \cap H_2|=16$, then, $H_1 \cap H_2$ is normal in both $H_1$ and $H_2$.

This means, $H_1 \cup H_2 \subseteq N(H_1 \cap H_2)$.

\begin{align*} |N(H_1 \cap H_2)|& \ge|H_1 \cup H_2|=|H_1|+|H_2|-|H_1 \cap H_2|\\&\ge32+32-16=48\\|N(H_1\cap H_2)|&\ge48\end{align*}

But, $N(H_1 \cap H_2)$ is a subgroup of the group of order $160$ and by Lagrange's theorem, we have, $$|N(H_1 \cap H_2)| \mid 160$$

Again, the only divisors of $160$ greater than 48 are $80$ and $160$.

If $|N(H_1 \cap H_2)|=80$, then $N(H_1 \cap H_2)$ is a index 2 subgroup and hence normal.

If $|N(H_1 \cap H_2)|=160$, then $H_1 \cap H_2$ is itself normal.

*Alternatively, @mk points out, this case can be solved even easily, since, the cardinality of the normalizer is bigger than $2^5$,(as it contains $H_1$ as its subgroup) and, as $2^5$ divides it, we must have that the normalizer is the whole group! *

Case 2: $|H_1 \cap H_2|=8$

We have that $n_5=2^4$.

We claim that all Sylow $2$-subgroups must intersect in a group of order $8$.

Suppose there were $3$ distinct Sylow $2$-Subgroups, they will exhaust $93$ non-trivial elements leaving no room for the remaining $2$ subgroups.

Suppose there were $2$ distinct Sylow $2$-Subgroups, and even if remaining three subgroups intersect in a group of order $8$, they are going to have too many elements.($24 \times 3+7+31 \times 2$=$141$ non-identity elements)

Suppose there is a unique distinct group, even then there will be excess of elements.$24\times 4 +7+31=134$ non-identity elements.

Now, the only possibility we left open will also lead to a contradiction.

Because, in this case we require, $127$ elements.

So, a simple counting helps!

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Why is $H_1 \cap H_2$ normal in $H_1$ and $H_2$? –  Mikko Korhonen Jan 27 '12 at 22:22
If you assume you have more than one 2-Sylow subgroup, then actually this will imply that they are self-normalizing, so clearly the intersection of such two will not be normal in either of them. If the intersection had been normal in each of them, the proof could have been completed by simply noting that then the normalizer contains two distinct maximal subgroups and would thus be the entire group. –  Tobias Kildetoft Jan 27 '12 at 22:31
Thanks, Kannappan :) –  group Jan 28 '12 at 1:17