## Exhibit all of the Sylow 3-subgroups of a special linear group

Exhibit all Sylow 3-subgroups of .

By this previous exercise, . We also know that . Thus . By Sylow’s Theorem, the number of Sylow 3-subgroups is congruent to 1 mod 3 and divides 8. Thus .

Let , with .

Note that and . If , then the entries of must satisfy the following four equations. Note that for all .

Suppose . Then equations (1) and (4) give . Now , but since otherwise . Thus . In either case, we generate the Sylow 3-subgroup

Now suppose . Then equations (1) and (4) give . Now , but since otherwise . Thus . In either case, we generate the Sylow 3-subgroup

Henceforth we may assume that and are not zero.

Suppose .

- If , then equation (1) gives us . Clearly then , as otherwise . Moreover, equation (2) gives . Since , we have . Thus generates the Sylow 3-subgroup
- If , then equation (1) gives us . Since , we have . Then equation (3) gives , a contradiction.
- If , then equation (1) gives us . Then , as otherwise . If , then , so that . Then equation (2) gives us , a contradiction. Thus , so that and we have . Thus generates the Sylow 3-subgroup

Suppose .

- If , then equation (1) gives us , so that . If , then and equation (2) gives us , a contradiction. Thus and ; in this case generates .
- If , then equation (1) gives us . Since , we have , so that . But then equation (1) gives , a contradiction.
- If , then equation (1) gives us . Then , as otherwise . If , then , so that . Then equation (3) gives , a contradiction. Thus , so that , and generates .

Thus , and we have exhausted the Sylow 3-subgroups of .

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## Comments

At the top it should say the number of Sylow 3 subgroups is congruent to 1 mod 3 and divides 8, so it’s 1 or 4.

Thanks!