1860 United States presidential election in Minnesota

The 1860 United States presidential election in Minnesota took place on November 6, 1860, as part of the 1860 United States presidential election. Minnesota voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.

Minnesota voted in its first ever United States presidential election, having been admitted as the 32nd state on May 11, 1858. The state was won by Illinois Representative Abraham Lincoln (Republican Party (United States)), running with Senator Hannibal Hamlin, with 57.23% of the popular vote, against Senator Stephen A. Douglas (D–Illinois), running with 41st Governor of Georgia Herschel V. Johnson, with 43.97% of the popular vote.

With 63.53% of the popular vote, Lincoln's victory within the state would be his second strongest victory in terms of percentage in the popular vote in the 1860 election after Vermont.^[1]

Results

1860 United States presidential election in Minnesota^[2]
Party		Candidate	Votes	%
	Republican	Abraham Lincoln	22,069	63.53%
	Democratic	Stephen A. Douglas	11,920	34.31%
	Southern Democratic	John C. Breckinridge	748	2.15%
	Constitutional Union	John Bell	50	0.14%
Total votes			34,787	100%

References

^ "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
^ "1860 Presidential Election Results Minnesota". {{cite web}}: Cite has empty unknown parameter: |1= (help)

[1] "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.

[2] "1860 Presidential Election Results Minnesota". {{cite web}}: Cite has empty unknown parameter: |1= (help)

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