Given a group , a subset is a called a *subgroup* if is itself a group under the same operation as . This relationship between and is denoted , and we write if is a proper subset of and we say is a *proper subgroup* of . Going through the “usual” process of verifying is a group isn’t necessary to see that a subset is a subgroup though, as some of these properties will be inherited from . For instance, associativity of the group operation doesn’t need to be checked. If the operation was *not* associative in , then there would be elements of for which the operation wasn’t associative and wouldn’t be a group. The only three things we need to check for to be a subgroup are that it contains the identity (which we’ll just call 1), that it’s closed under the group operation (i.e., ) and closed under inverses ().

In fact we don’t even have to show all of those properties. The *subgroup criterion* (as Dummit & Foote call it), aka the two-step subgroup test (Gallian) says that all we need to do is show that is non-empty, and for every that . Obviously these hold if , so we need to show that if these properties hold then . Letting , then we have that . Now let and we have . Finally, since we’ve shown that is closed under inversion, we have that for and so is a subgroup of .

As an example, let be under vector addition. We will show that the set of all points lying on the line form a subgroup of . Let . This is clearly a non-empty set as . Now let where and . Note the inverse of is . Adding our vectors together we have and so is a subgroup of .

If is any set, we denote by the set of all bijective functions (normally just called permutations) of . We claim that is a group under function composition. Note that function composition is associative as for any and that . There is an identity element given by the map , and that for any we have an inverse by mapping back to : by assumption is bijective so there are no problems with having multiple things map to under , or mapping multiple things to under .

In the case we generally write in place of . We also generally write these permutations in *cycle notation*. A *cycle* is a parenthesized list of distinct numbers where the first number maps to the second, the second to the third, the third to the fourth, and so on. The last number in the list maps back to the first number, hence the name “cycle.” For example, (1 4 5 3 2) means , , , and finally . In the cycle decomposition of a function we list all such cycles. If we see (1 4 5 3 2)(6 7)(8), for instance, we have the cycle previously discussed, but in addition we have and by the second cycle, and the third cycle tells us (elements that map to themselves are called *fixed points*). The convention is not to list any fixed points in the cycle decomposition though, with the understanding that any any points not appearing in the decomposition map to themselves, so the above permutation would normally be written (1 4 5 3 2)(6 7). Of course, this permutation is equivalently given by (7 6)(3 2 1 4 5). By convention though, we list the cycle containing 1 first, the cycle containing the lowest number not in the first cycle next, and so forth. In a given cycle we will normally list the least element first.

Now let be set of all permutations of that fix the elements 3 and 7. We will show that . Obviously . Let . Note that if 3 and 7 are fixed by , they will be fixed by as well. Applying after we still keep 3 and 7 fixed. This gives us that .

[…] group, some index set, and suppose for every . We will show that by applying the subgroup test previously described. First note that the identity is in since each means the identity is in each . Now let . […]

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