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My question is suppose A is a subgroup of the group G. If e_G is the identity element of G and e_A is the identity element of A, how can I prove that e_G=e_A.

I was thinking of doing this by contradiction, but I am not too sure if this is the best approach.

Would a better choice be to use the identity property of groups?

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Hint: See if you can prove that $e_Ge_A=e_Ae_A$. How can you then conclude that $e_G=e_A$?

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    Then should I start assuming e_G=e_A? Would this then involve the inverse property of groups? – 2017-02-03
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    No, that's what you want to prove, so you can't assume it. Instead you'll just have to use the definition of $e_G$ being the identity of $G$ and $e_A$ being the identity of $A$. You will need to use inverses for the second step (getting from $e_Ge_A=e_Ae_A$ to $e_G=e_A$). – 2017-02-03
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    okay, so I should pick some element in G say b so I can say b*e_G=b=e_G*b – 2017-02-03
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    Oh okay, I think I understand somewhat. Well I would then use the inverse and rearranging the elements around. – 2017-02-03
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    Can you prove that a group has one and only one element satisfying the equation $x^2=x?$ – 2017-02-03
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    @bof I am not sure I know how too...I just started learning about groups. I would think for example that only 1 works here – 2017-02-03
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    @EricWofsey Can I start with something like since e_A is an identity on A and it is in A then e_A * e_A = e_A? – 2017-02-04
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    Yep, that's a good start! – 2017-02-04