1889 Chicago mayoral election explained

Election Name:1889 Chicago mayoral election
Type:presidential
Ongoing:no
Previous Year:1887
Next Year:1891
Election Date:April 2, 1889
Image1:DeWitt Clinton Cregier old portrait (1).jpg
Nominee1:DeWitt Clinton Cregier
Party1:Democratic Party (United States)
Popular Vote1:57,340
Percentage1:54.93%
Nominee2:John A. Roche
Party2:Republican Party (United States)
Popular Vote2:46,328
Percentage2:44.38%
Mayor
Before Election:John A. Roche
Before Party:Republican Party (United States)
After Election:DeWitt Clinton Cregier
After Party:Democratic Party (United States)

In the Chicago mayoral election of 1889, Democrat DeWitt Clinton Cregier defeated incumbent Republican John A. Roche, winning a majority of the vote and a margin of victory in excess of ten percent.

The election was held on April 2, 1889.[1]

Campaign

Cregier backed strongly by trade unions.[2] John Peter Altgeld threw his backing behind Cregier's candidacy.[2]

Results

Creiger received 76.86% of the Polish-American vote, while Roche received 23.07%.[3]

Notes and References

  1. Book: Chicago: Its History and Its Builders, a Century of Marvelous Growth. 335 . S. J. Clarke publishing Company . Currey. Josiah Seymour. 1912.
  2. Book: Kersten . Andrew E. . Clarence Darrow: American Iconoclast . 2011 . Macmillan . 978-0-8090-9486-8 . 46 . 2 June 2020 . en.
  3. Kantowicz . Edward . The Emergence of the Polish-Democratic Vote in Chicago . Polish American Studies . 1972 . 29 . 1/2 . 67–80 . 20147849 .