The 1864 United States presidential election in Indiana took place on November 8, 1864, as part of the 1864 United States presidential election. Indiana voters chose 13 representatives, or electors, to the Electoral College, who voted for president and vice president.[1]
 | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
 County Results | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
.jpg)
Indiana was won by the National Union candidate, incumbent Republican President Abraham Lincoln of Illinois and his running mate former Senator and Military Governor of Tennessee Andrew Johnson. They defeated the Democratic candidate, 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton of Ohio. Lincoln won the state by a margin of 7.2%.[1]
Results
| 1864 United States presidential election in Indiana[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| National Union | Abraham Lincoln of Illinois | Andrew Johnson of Tennessee | 150,422 | 53.60% | 13 | 100.00% | ||
| Democratic | George B. McClellan of New Jersey | George H. Pendleton of Ohio | 130,233 | 46.40% | 0 | 0.00% | ||
| Total | 280,655 | 100.00% | 13 | 100.00% | ||||