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For an object $X$ in the infinity category of pointed space $S_{*}$, if it has an $E_{\infty}$ grouplike structure, then it give rises to a unique infinite delooping $BX$, which is a connective spectrum and has the property that $X\cong\Omega^{\infty}BX$. Moreover, there is an equivalence between connective spectrums and $E_{\infty}$ grouplike spaces where for a connective spectrum $E$, it corresponds to its zero part $\Omega^{\infty}(E)=E[0]$, which has the 'group structure' behaving like the concatenation of paths in the loop space.

I just get curious, if a pointed space $X\in S_{*}$ admits a $E_{\infty}$ group structure, is its infinite delooping $BX$ exactly its infinite suspension $\Sigma^{\infty}X$?

Remark: thanks for the comments by Ulrik Buchholtz, I realize my mistake, the unit map $X\rightarrow \Omega^{\infty}\Sigma^{\infty}X$ is the underlying space of the $E_{\infty}$ group $\Sigma^{\infty}X$ (under the equivalence between connective spectrums and $E_{\infty}$ group spaces), the simplest contradiction is given by the comment: $\Omega^{\infty}(\Sigma^{\infty}S_{0})$ is not a two-element space.

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    $\begingroup$ No, it gives you the infinite delooping of the free symmetric infinity group on the underlying space. It's just like taking the free group on the underlying set of a group never gives you back the group you started with. (To begin with, you got a new neutral element.) As another example, consider the 0-sphere with one of its two group structures. The infinite suspension gives you the sphere spectrum, i.e., representing the free symmetric infinity group on one generator, and this is not a 2-element group. $\endgroup$
    – Ulrik Buchholtz
    Commented Sep 3 at 7:40
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    $\begingroup$ @UlrikBuchholtz: Maybe convert that to an answer? Having an essentially-complete answer in comments tends to put other people off from writing the same thing in an answer, so the question gets stuck unanswered… $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Sep 3 at 15:50
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    $\begingroup$ IMHO this is a case where you can’t see the wood for the trees. If you strip down the question from all the unnecessary higher categorical decoration it amounts to asking whether any infinite loop space is the infinite loop space of its infinite suspension. This is an advanced but not quite a research level question on classical algebraic topology whose answer is well known. All this reminds me of mathoverflow.net/a/83307/12166 $\endgroup$
    – Fernando Muro
    Commented Sep 3 at 17:08
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    $\begingroup$ @FernandoMuro I don't see it that way. I see it as someone asking a very common question for newcomers to the field, and doing so with the language they have available. Nothing here is being overcomplicated by categorical decoration, beyond maybe the first sentence where the term "infinity category" was used. $\endgroup$
    – Jonathan Beardsley
    Commented Sep 3 at 21:13
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    $\begingroup$ Although I do agree that this is not quite a research level question. But on the other hand, lots of questions are on MathOverflow which are more of an "important for young researchers to know and be able to search for" nature. $\endgroup$
    – Jonathan Beardsley
    Commented Sep 3 at 21:15

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(I'm converting my comment to an answer:)

No, it gives you the infinite delooping of the free symmetric infinity group on the underlying space. It's just like taking the free group on the underlying set of a group never gives you back the group you started with. (To begin with, you got a new neutral element.) As another example, consider the $0$-sphere with one of its two group structures. The infinite suspension gives you the sphere spectrum, i.e., representing the free symmetric infinity group on one generator, and this is not a $2$-element group.

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