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if $|G|=4$ with no element of order $4$ show non-identity element has no order of 2

$|G|=4$ so has 4 elements $$ G=\{e,a,b,c \}$$

No non identy element has order of 4 so for any non idenitty element $g\in G;|g|\neq 4 \iff g^4\neq e $

We want to show for all non-Identity elment of has no order of $2$

suppose exists an element $g$ either $a,b,c$ where $|g|=2$ that means that it is its own inveres so $g^{-1}=g$

setting $a=g=a^{-1}$ it should be that $ab=c$ and $ac=b$ since a is a non-identity

not sure how to reach a contradiction. I think it has to do something. with divisiblity? and also this question might be anwered somewhere here

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    Is this formulation correct : "non-idendity element has no order of $2$" ? A non-idendity element in $\mathbb Z_4$ must have order $2$ or $4$ ... – 2017-01-30
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    If you are allowed to use that there are only two groups with order $4$, the cyclic group and the Klein-four-group, it is easy. – 2017-01-30
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    Assuming that you are new to Groups, I am giving a proof which is quiet elementary. If $G$ is a finite group, then the order of an element must divide the order of the Group. Here $|G|=4$, so the elements of $G$ can have orders $1$,$2$ or $4$. Since your group does not have an element of order $4$, this leaves us with elements of order $1$ and $2$. Now $1$ must be the order of the identity element. And since you are asking for non-identity elements we are left with elements of order $2$. Hence all non-identity elements must have order $2$ – 2017-01-30
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    "The order of $g$ is not $4$" does not mean "$g^4 \neq e$". It means "The smallest positive integer $n$ such that $g^n = e$ (if it exists) is not $4$". Specifically, it doesn't exclude the possibility that the order of $g$ is $2$. – 2017-01-30

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A group of order $4$ is always $\mathbb Z/4\mathbb Z$ or $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ (up to isomorphism). Since it has no element of order $4$, it must be $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$.

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    Is not "always", any group of order 4 is isomorphic to the groups u listed. – 2017-01-30
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    @Shobhit: You right ! But since two isomorphic group are considered to be the "same", I don't think that there is ambiguity ;-). But ok, I precise it. – 2017-01-30