1886 Georgia gubernatorial election
Georgian gubernational election
From Wikipedia, the free encyclopedia
The 1886 Georgia gubernatorial election was held on October 6, 1886, in order to elect the Governor of Georgia. Democratic nominee, former United States Senator from Georgia and candidate for Governor in the 1868 election John B. Gordon ran unopposed and thus won the election.[1]
 | |||||||||||||||||
October 6, 1886
| |||||||||||||||||
| |||||||||||||||||
 Results by County: Gordon: >90% | |||||||||||||||||
| |||||||||||||||||
General election
On election day, October 6, 1886, Democratic nominee John B. Gordon won the election with 99.20% of the vote, thereby holding Democratic control over the office of Governor. Gordon was sworn in as the 53rd Governor of Georgia on November 9, 1886.[2]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | John B. Gordon | 101,159 | 99.20 | |
| Others | 815 | 0.80 | ||
| Total votes | 101,974 | 100.00 | ||
| Democratic hold | ||||