1886 Georgia gubernatorial election
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| Elections in Georgia |
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The 1886 Georgia gubernatorial election was held on 6 October 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]
General election
On election day, 6 October 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 9 November 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 | ||||
References
- ^ "Gov. John Brown Gordon". National Governors Association. Retrieved 4 January 2024.
- ^ "GA Governor". ourcampaigns.com. 1 September 2005. Retrieved 4 January 2024.
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