# Show that a group of order $90$ is solvable

Let $$G$$ be a group of order $$90$$. Show that $$G$$ is solvable.

Given that $$90=2 \cdot 5 \cdot 3^2$$ is of the type $$pqr^2$$, with $$p, q,$$ and $$r$$ primes, this case is not straightforward.

My attempt to find a solution: The number of Sylow 5-subgroups of $$G$$, $$n_5$$, is either $$1$$ or $$6$$.

Case 1: $$n_5=1$$. In the first case, the only subgroup (say $$K_5$$) is normal. Then, $$G$$ will be solvable if both $$K_5$$ and $$G/K_5$$ are solvable. $$K_5$$ is solvable as it has order 5. $$G/K_5$$ is also solvable because its order is $$90/5=18=2\cdot 3^2$$ (of the type $$pq^2$$); therefore $$G$$ is solvable.

Case 2: $$n_5=6$$. There are 6 Sylow 5-subgroups which are not normal and I don't know how to continue from here.

Similar arguments with $$r=3$$ made me get stuck as well when $$n_3=10$$.

• First of all, $90/5=18=2*3*3...$ – Mindlack Jan 7 at 0:01
• Thanks for pointing that out! Corrected. – pendermath Jan 7 at 0:04

From what you've done the only case to worry about is when there are 10 Sylow 3-subgroups and 6 Sylow 5-subgroups.

You can conclude that the Sylow 3-subgroups not all disjoint. Otherwise you would have 8 non-identity elements from each of the 10 Sylow 3-subgroups, leaving only 10 elements which would not be compatible with having 6 Sylow 5-subgroups.

Consider a nontrivial element $$x$$ in some non-trivial intersection of two Sylow 3-subgroups. Let $$H$$ denote the centralizer of $$x$$. Since groups of order 9 are abelian, $$H$$ contains multiple Sylow 3-subgroups. The possible values of $$|H|$$ are thus either 18, 45, or 90.

If $$|H|= 45$$ then the centralizer of $$x$$ is normal (being index 2), and we are reduced to showing that groups of order 45 are solvable.

If $$|H|= 90$$, then $$x$$ is central, so we are reduced to showing that $$G/\langle x\rangle$$, which is order 30, is solvable. This is an exercise left to the reader.

If $$|H|= 18$$, then $$H$$ might not be normal, but consider the left action of $$G$$ on left cosets of $$H$$. This yields a homomorphism $$\phi: G \to S_5$$. The kernel of this map is a normal subgroup of $$G$$ that is contained in $$H$$, since $$H$$ is the stabilizer of the identity coset. Moreover, the order of the kernel of $$\phi$$ is a multiple of 3. This kernel is order either 3, 6, 9, or 18, with the quotient group order either 30, 15, 10, or 5. We are thus reduced to checking that all groups of such orders are solvable.

• Thanks for the detailed response. You lost me in the last case (|H|=18), hopefully there is another way to prove this case. Anyways, I will take my time to think about it. Thanks! – pendermath Jan 7 at 0:47
• Minor thing: your element in the intersection switched from being called $x$ to $g$ and back again. – user3482749 Jan 7 at 12:35
• Yes, that drove me mad for a while, @user3482749 – pendermath Jan 7 at 13:00
• thanks for the notice on the inconsistent notation, fixed – Rolf Hoyer Jan 8 at 1:35

There is an injection $$f$$ of $$G$$ into $$S_{90}$$ via the permutation action of $$G$$ on itself by left multiplication. An element of order 2 maps to a disjoint product of 45 transpositions, which is an odd permutation, so the image of $$f$$ is not contained in $$A_{90}$$. Since $$S_{90}/A_{90}$$ has order 2, composing that quotient map on $$f$$ yields a surjection of $$G$$ on to the group of order 2. The kernel is a normal subgroup of order 45, so $$G$$ is not normal.

Similarly any group of order greater than 2 whose order is divisible 2 but not 4 has a nontrivial normal subgroup of index 2.

• This is a slick argument! I had a feeling my answer wasn't the simplest. – Rolf Hoyer Jan 8 at 2:25