| Elections in Michigan |
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The 1860 United States presidential election in Michigan took place on November 6, 1860, as part of the 1860 United States presidential election. Voters chose six representatives, or electors, to the Electoral College, who voted for president and vice president.[1]
Michigan was won by the Republican nominee, Illinois Representative Abraham Lincoln and his running mate Senator Hannibal Hamlin of Maine. They defeated the Democratic nominee, Senator Stephen A. Douglas of Illinois and his running mate with 41st Governor of Georgia Herschel V. Johnson.[1] Lincoln won the state by a margin of 15.24%.
The 1860 presidential election in Michigan began a trend in which the state would vote the same as Pennsylvania, as the two states would vote in lockstep with each other on all but three occasions since Lincoln’s victory - 1932, 1940, and 1976.
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Abraham Lincoln | 88,450 | 57.23% | |
| Democratic | Stephen A. Douglas | 64,889 | 41.99% | |
| Southern Democratic | John C. Breckinridge | 805 | 0.52% | |
| Constitutional Union | John Bell | 405 | 0.26% | |
| Total votes | 154,549 | 100% | ||
