# Groups of orders 147

### also 273, 100 and 260

There are six distinct (non-isomorphic) groups of order 147.

If we try to list them, we find five:

1. C7 × C7 × C3
2. C7⋊C3 × C7
3. C49 × C3
4. C49⋊C3
The last of these is the pullback of C7⋊C3 and C7⋊C3 by C3. We will, for convenience, write it as (C7×C7)⋊C3.

It may not be obvious where the sixth group of order 147 is. In fact, the last group listed above listed is not well-defined: there are two different groups which can be described as (C7×C7)⋊C3, or pullbacks of C7⋊C3 and C7⋊C3 by C3. The purpose of this page is to show how this happens.

## The groups (C7×C7)⋊C3

Consider the group (ℤ7,+), the integers modulo 7 under addition. It is isomorphic with C7. It has six automorphisms (forming C6), corresponding to multiplying its elements by 1,2,3,4,5 and 6. Of these, two, multiplication by 2 and by 4, have period three. Mutiplication by 2 does the permutation (1 2 4)(3 6 5), and multiplication by 4 does the permutation (1 4 2)(3 4 6). These two permutations, and the identity, form the C3 which acts on C7 in C7⋊C3.

It is significant that these two permutations are distinguishable. If we find an abstract C7 lying around, we won't know which of its elements to label as "1", any of the six non-identity elements will do. But whichever one we choose to call "1", if someone then does a period-3 automorphism on it, we will be able to tell whether they have done (1 2 4)(3 6 5), or (1 4 2)(3 4 6). The former permutes each element x to x+x, the latter, to x+x+x+x. The two period-3 automorphisms are distinguishable.

Now, if we are just considering C7⋊C3, a C7 permuted by a C3, this doesn't matter. The C3 has two non-identity elements, one doing (1 2 4)(3 6 5) and the other doing (1 4 2)(3 4 6). So all groups C7⋊C3 are isomorphic. But when we have (C7×C7)⋊C3, two C7s permuted by the same C3, there are two distinguishable ways of building the group. In one case, one element of the C3s permutes both C7s "forwards", (1 2 4)(3 6 5), and the other permutes them both backwards, (1 4 2)(3 4 6). In the other case, each non-identity element of the C3 permutes one C7 forwards and the other backwards.

Thus we have found the six groups of order 147:

1. C7 × C7 × C3
2. C7⋊C3 × C7
3. C49 × C3
4. C49⋊C3
5. (C7×C7)⋊C3    both C7s the same way
6. (C7×C7)⋊C3    the two C7s opposite ways

## Groups of order 273

Something similar happens with groups of order 273. There are five distinct groups:

1. C13 × C7 × C3
2. C13⋊C3 × C7
3. C7⋊C3 × C13
4. (C13×C7)⋊C3        the C7 and the C13 go "the same way"
5. (C13×C7)⋊C3        the C7 and the C13 go "opposite ways"

Here it is more arbitrary how we define "the same way". For the C7 the period-3 permutations are (1 2 4)(3 6 5) and (1 4 2)(3 4 6), for C13 they are (1 3 9)(2 6 5)(4 12 10)(7 8 11) and (1 9 3)(2 5 6)(4 10 12)(7 11 8). Which we choose to label "forwards" and which "backwards" is arbitrary; but whichever labelling we choose, we can decide whether by our labelling a group (C13×C7)⋊C3 has the C7 and the C13 going the same way or opposite ways when permuted by an element of the C3.

## Groups of order 100

The smallest order for which this "twinning" effect happens is 100. But here things are more complicated. The 16 groups of order 100 are:

• Sylow subgroups C5×C5, C2×C2
1. C5 × C5 × C2 × C2
2. C5⋊C2 × C5 × C2
3. C5⋊C2 × C5⋊C2
4. pullback of C5⋊C2 and C5⋊C2 by C2, × C2
• Sylow subgroups C25, C2×C2
1. C25, C2×C2
2. C25⋊C2 × C2
• Sylow subgroups C5×C5, C4
1. C5 × C5 × C4
2. C5⋊C2C4 × C5
3. C5⋊C4C4 × C5
4. pullback of C5⋊C2C4 and C5⋊C2C4 by C4
5. pullback of C5⋊C4C4 and C5⋊C2C4 by C4
6. pullback of C5⋊C4C4 and C5⋊C4C4 by C4        The C4 turns the C5 the same way
7. pullback of C5⋊C4C4 and C5⋊C4C4 by C4        The C4 turns the C5s opposite ways
• Sylow subgroups C25, C4
1. C25 × C4
2. C25⋊C2C4
3. C25⋊C4C4

## Groups of order 260

There are 15 groups of order 260:

• Sylow subgroups C13, C5, C2×C2
1. C13 × C5 × C2 × C2
2. C5⋊C2 × C13 × C2
3. C13⋊C2 × C5 × C2
4. C13⋊C2 × C5⋊C2
5. pullback of C13⋊C2 and C5⋊C2 by C2, × C2
• Sylow subgroups C13, C5, C4
1. C13 × C5 × C4
2. C5⋊C2C4 × C13
3. C5⋊C4C4 × C13
4. C13⋊C2C4 × C5
5. C13⋊C4C4 × C5
6. pullback of C13⋊C2C4 and C5⋊C2C4 by C4
7. pullback of C13⋊C4C4 and C5⋊C2C4 by C4
8. pullback of C13⋊C2C4 and C5⋊C4C4 by C4
9. pullback of C13⋊C4C4 and C5⋊C4C4 by C4        The C4 turns the C5 and the C13 "the same way"
10. pullback of C13⋊C4C4 and C5⋊C4C4 by C4        The C4 turns the C5 and the C13 "opposite ways"

## Groups of order 48

We might think that the groups of order 48 would also show this "twinning", with two non-isomorphic groups (C22,C22)⋊C3, both pullbacks of A4 and A4 by C3. This would be wrong. If we have a period-3 automorphism of C7, we can tell whether it goes "forwards" or "backwards". But C22⋊C3, ≅ A4, is not like that. The two period-3 automorphisms of C22 are indistinguishable.