1824 United States presidential election in Georgia

Summary

The 1824 United States presidential election in Georgia took place on November 1, 1824, as part of the 1824 United States presidential election. The Georgia General Assembly chose 9 representatives, or electors to the Electoral College, who voted for President and Vice President.

1824 United States presidential election in Georgia

← 1820 October 26 – December 2, 1824 1828 →
 
Nominee William H. Crawford Andrew Jackson
Party Democratic-Republican Democratic-Republican
Alliance Crawford party Clark party
Home state Georgia Tennessee
Running mate Nathaniel Macon John C. Calhoun
Electoral vote 9 0
Popular vote 121 45
Percentage 72.89% 27.11%

President before election

James Monroe
Democratic-Republican

Elected President

John Quincy Adams
Democratic-Republican

During this time, political divisions in the state fell along the lines of personal support for outstanding leaders in their struggle for power. Many of these factions were usually held together through personal friendships and family associations. The two factions at the time were the Clark faction, followers of Ex-Governor John Clark, and the Crawford faction, followers of Secretary of the Treasury William H. Crawford.[1][2]

As both houses of the assembly were controlled by the Crawford party, their slate won by a margin of 121 to 45 against the Clark party which nominated Andrew Jackson.[2]

Results edit

1824 United States presidential election in Georgia[2]
Party Candidate Votes Percentage Electoral votes
Democratic-Republican William H. Crawford 121 72.89% 9
Democratic-Republican Andrew Jackson 45 27.11% 0
Totals 166 100% 9

References edit

  1. ^ Murray, Paul (1945). "Party Organization in Georgia Politics 1825-1853". The Georgia Historical Quarterly. 29 (4): 195–196. JSTOR 40576991 – via JSTOR.
  2. ^ a b c Lamplugh, George (2015). Rancorous Enmities and Blind Partialities: Factions and Parties in Georgia, 1807–1845. McFarland. p. 107. ISBN 9780761865872.