Idaho gubernatorial election
1890 Idaho gubernatorial election![](//upload.wikimedia.org/wikipedia/commons/thumb/a/a4/Flag_of_Idaho.svg/50px-Flag_of_Idaho.svg.png)
|
|
| | | Nominee | George L. Shoup | Benjamin Wilson | | Party | Republican | Democratic | Popular vote | 10,262 | 7,948 | Percentage | 56.35% | 43.65% | |
![](//upload.wikimedia.org/wikipedia/commons/thumb/2/27/1890_Idaho_gubernatorial_election_results_map_by_county.svg/160px-1890_Idaho_gubernatorial_election_results_map_by_county.svg.png) County results Shoup: 50–60% 60–70% Wilson: 50–60% |
Governor before election George L. Shoup (Territorial) Republican | Elected Governor George L. Shoup Republican | |
Elections in Idaho |
---|
![](//upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Seal_of_Idaho.svg/150px-Seal_of_Idaho.svg.png) |
|
|
|
|
|
|
The 1890 Idaho gubernatorial election was held on 1 October 1890, in order to elect the first Governor of Idaho upon Idaho acquiring statehood in July 1890. Incumbent Republican Governor of the Idaho Territory George L. Shoup defeated Democratic nominee Benjamin Wilson.[1]
General election
On election day, 1 October 1890, Republican nominee George L. Shoup won the election by a margin of 2,314 votes against his opponent Democratic nominee Benjamin Wilson, thereby retaining Republican control over the new office of Governor. Shoup was sworn in as the first Governor of the new state of Idaho on 8 December 1890.[2]
Results
Idaho gubernatorial election, 1890 Party | Candidate | Votes | % |
| Republican | George L. Shoup (incumbent[a]) | 10,262 | 56.35% |
| Democratic | Benjamin Wilson | 7,948 | 43.65% |
Total votes | 18,210 | 100.00% |
| Republican hold |
Notes
References
- ^ "George L. Shoup". National Governors Association. Retrieved May 12, 2023.
- ^ "ID Governor". ourcampaigns.com. September 26, 2005. Retrieved May 12, 2023.