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FBI place first ever woman on 'most wanted' list

Joanne Chesimard, who fled to Cuba after being convicted of killing a state trooper in 1973, is the first woman ever to make the list.

A poster with photographs of Joanne Chesimard. (Julio Cortez/AP/Press Association Images)

A WOMAN WHO killed a police officer 40 years ago and later fled to Cuba has been placed on the FBI’s list of “most wanted terrorists,” US officials said Thursday.

Law enforcement officials said Joanne Chesimard is the first woman ever to make the list, and one of only two to make the list accused of crimes carried out on US soil.

“Joanne Chesimard is a domestic terrorist who murdered a law enforcement officer execution-style,” said FBI agent Aaron Ford, in a statement issued four decades to the day after the death of the New Jersey State Police trooper.

“We want the public to know that we will not rest until this fugitive is brought to justice,” he said.

Black Liberation Army

US authorities say that on May 2, 1973, Chesimard, a member of the radical Black Liberation Army group, along with two accomplices, fatally shot the highway police officer during a traffic stop.

She was convicted of first-degree murder, armed robbery, and other crimes in 1977 and sentenced to life in prison.

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Chesimard escaped prison less than two years later, and lived underground before surfacing in 1984 in Cuba, where she had sought political asylum, the FBI said.

The FBI is offering a reward of up to $1 million for information leading to the capture of Chesimard, while the state of New Jersey is offering a separate $1 million reward.

- © AFP, 2013

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