1886 Georgia gubernatorial election
Appearance
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Results by County: Gordon: >90% | |||||||||||||||||
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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
[edit]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
[edit]Party | Candidate | Votes | % | |
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Democratic | John B. Gordon | 101,159 | 99.20 | |
Others | 815 | 0.80 | ||
Total votes | 101,974 | 100.00 | ||
Democratic hold |
References
[edit]- ^ "Gov. John Brown Gordon". National Governors Association. Retrieved 4 January 2024.
- ^ "GA Governor". ourcampaigns.com. 1 September 2005. Retrieved 4 January 2024.
Categories:
- Georgia (U.S. state) gubernatorial elections
- 1886 United States gubernatorial elections
- 1886 Georgia (U.S. state) elections
- October 1886 events
- Single-candidate elections
- 1886 in Georgia (U.S. state)
- 1880s in Georgia (U.S. state)
- 1880s Georgia (U.S. state) elections
- 1886 elections in North America
- 1886 elections
- 1886 elections in the United States
- United States gubernatorial elections in the 1880s
- Government of Georgia (U.S. state)