Base case when applying induction in group theory There are always two kinds of induction in general:

*

*Assume that the result holds for $n-1$

*Assume the the result holds for integers $<n$
For the former one, we only need to check one base case; for the latter one, we always need to check two base cases.
I’m learning group theory. Induction is a very useful tool that is always employed. We always prove by induction on group order $|G|$ and assume that results hold for groups of order $<|G|$. But I find that we always check only one base, namely $|G|=1$! Maybe I misunderstood something, but I really don’t remember seeing any proof checking more than one base case.
So what is the point? Could you give me some ideas? Any help would be appreciated!
Possible examples:

*

*Index of maximal proper subgroup of a solvable group


*https://math.stackexchange.com/a/2391609/792898
 A: Here are some reasons why one might want to check more than one base case:

*

*the argument used in the inductive step cannot be applied for small $n$, hence the proof writer checks those cases separately in the base cases

*the inductive step (proof for $n+1$) explicitly refers to the cases $n$ and $n-1$ (sometimes also $n-2$ etc.)

*the writer might want to get a better feeling for what needs to be shown, and therefore checks the statement manually for more $n$ than actually needed.

In general, it is therefore a good idea to start (on a separate sheet of paper) with the inductive step. Afterwards, you know which base cases need to be checked.
The same applies for reading a proof: If you wonder why more than one base case was treated, take a look at the inductive step. The answer will most likely be hidden there.
A: Here is an induction that requires more than one base case. Say we have two stamps, one 5 cent and the other 3 cents. I claim that any number $n \geq 8$ can be made using just these two stamps. First we have $5+3=8$ and we have $3 + 3+ 3=9$. From these two base cases we can construct the next number by either replacing a $5$ with two $3$s or replace three $3$s with two $5$s. to complete the induction. Note that we do need the two base cases to ensure there will always be enough $5$s or $3$s to make the replacement for the inductive step.
