A week ago, I published a Hasse diagram for the subgroups of the symmetries of the cube, which is also the symmetries of the octahedron and is isomorphic to S4. There were two problems with my diagram.
1) It was hard to read.
2) There are actually 30 subgroups of this group instead of 27.
Is my face red? Yes, it is. Gotta fix that diagram.
In the group we deal with today, the order is 60 and there are 59 subgroups. Any regular Hasse diagram will be a mess. I hit upon a new way to simplify such a diagram, and I present it here.
The numbers on the left are the orders of the subgroups.
The numbers in the squares or circles represent how many copies of that order exist that are isomorphic to each other.
Examples: At order 60, there is only 1 subgroup, and that is the group itself.
At order 12, there are 5 isomorphic copies, at order 10, there are 6 isomorphic copies, etc.
A red circle indicates the subgroup is normal in the group.
A blue square means the group isn't normal.
Red arrows can only start from red circles, indicating a subgroup that is normal in the entire group must be normal in any supergroup to which it belongs.
A blue arrow shows that a subgroup is not normal in the next supergroup above it.
A purple arrow indicates a subgroup is not normal in the entire group, but is normal in some supergroup above it, but not necessarily in every supergroup above it. This is not the definition of subnormal, so there should be a different word for it. I do not yet know that word, so I have coined normalish to describe this condition.
Take a look at level 3. Any subgroup of order 3 will be normal in a supergroup of order 6, hence the purple arrow, but in this case, the subgroups of order 3 are not normal in the subgroup of order 12, hence the blue arrow. Likewise, the subgroups of order 2 must be normal if they are contained in a supergroup of order 4, but in this group, they are not normal when contained supergroups of order 10 or 12.
Notice that there are divisors of 60 that are not orders of subgroups of this particular group. We have seen that the order of a subgroup H of a finite group G must divide the order of G. This is known as LaGrange's Theorem. The converse would be that if a divides b and b is the order of G, there must be a subgroup of order a in G. This example shows us the converse is not true.
Usually, when something is simplified, we have lost information, and that is true here. If each of the 59 subgroups got its own node in this graph, we could see exactly which subgroups of order 2 are contained in each of the subgroups of order 10. There are six subgroups of order 10, and each would have five blue arrows coming up from five of the fifteen order 2 subgroups. The diagram would turn into a tangled mass of multi-colored pasta, and while more information would be shown, it would be a challenge to decipher it.
This next week is going to be cleaning up previous diagrams, and seeing if someone more knowledgeable than myself knows the real word that is used where I am using normalish. It is a common occurrence in well-known groups of small order, someone must have noticed it before I did.
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