Adam Lallana voted England's Player of the Year
Adam Lallana has been voted by fans as England's 2016 Player of the Year in a season when many of the England international squad failed to shine following a dismal performance at Euro 2016 ending in a disastrous defeat against international minnows Iceland. Lallana won the vote from the England Supporters Club with 39% of votes ahead of Jamie Vardy with 12% and two-time previous winner Wayne Rooney on 8%. He expressed his delight at receiving the award from supporters ahead of his England teammates declaring that it is a huge honour.
In 2015 and 2014 Wayne Rooney won the award and in 2012 former England captain Steven Gerrard won the vote.
Lallana's performances for England
Since making his way through the International Under 18, 19 and 21 ranks he has secured a place as a regular for England.He featured in most of the dismal Euro 2016 performances under Roy Hodgson with the exception of the defeat against Iceland. However, his best performances have come later in the season, and he scored his first England goal in Sam Allardyce's only game in the 1-0 victory over Slovakia. He then went on to net against Scotland and Spain at Wembley.
His performances at Liverpool
Since joining Liverpool in 2014 he has reached 76 appearances and 16 goals.He has so far enjoyed a good season at club level and stated in an interview that he was pleased with his form at club level and hoped that this would transfer onto the international stage.
Lallana will be hoping his form can continue as England return to the qualifiers for the 2018 World Cup in March with a game against Lithuania at Wembley.
What does the future hold?
Having proved himself already this season with successive England managers and his own team manager Jurgen Klopp there is every likelihood that he will continue to be an influential player for England for many years to come. He has admitted that the year for England has been a disappointment but with a new manager in place in Gareth Southgate, he will be looking to improve.